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✓ 14 results · all verified · 13 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Sequences and Limits: Examples and Counterexamples

1 · Prerequisites

2 · Summary

This companion computes with the toolkit of sequences-and-limits. It opens with the reference null sequence $1 / (k + 1)$ of The sequence $1 / (k + 1)$ converges to $0$ , proved from density of $Q$ and the Archimedean property of $Q$ alone, and then spends it: a rational expression evaluated by the algebra of limits, an oscillating null sequence that separates the sign pattern from the size of the terms, the Archimedean divergence of $\sqrt{k}$ , and the successive differences $\sqrt{k + 1} - \sqrt{k}$ , which converge although both terms diverge. Tail invariance is worked out on the classical instance of changing the first hundred terms, and the two squeeze computations cover the standard root squeeze and a quotient that is not presented as a sum of two known limits.

The boundary cases are carried by counterexamples, one per load-bearing hypothesis: a null sequence times an unbounded one need not be null (boundedness cannot be dropped from A null sequence times a bounded sequence is null); the sum of two divergent sequences can converge, so the sum rule may not be read backwards; the quotient rule needs a nonzero limit in the denominator; and the squeeze theorem needs the two bounds to share their limit. Two further examples isolate the sharply true halves of the surrounding statements: subsequences of a null sequence inherit the limit, and the absolute value of the limit is recovered from the limit of the absolute values while the converse fails at every nonzero limit.

The page closes with the complete classification of the geometric sequence $r^{k}$ for every real ratio, from integer-power recursion and the Bernoulli and Archimedean estimates proved on roots-and-rational-powers: null for $∣ r ∣ < 1$ , constant for $r = 1$ , bounded but divergent at $r = - 1$ , growing to $+ \infty$ for $r > 1$ , and sign-alternating with $∣ r^{k} ∣ \to + \infty$ for $r < - 1$ , so that $(r^{k})$ converges exactly for $- 1 < r \leq 1$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The sequence $1 / (k + 1)$ converges to $0$

Example

The sequence of reals $x_{k} : = (k + 1)^{- 1}$ , with $k + 1 = σ (k)$ the successor of $k$ read in $R$ through the canonical-natural embedding (Canonical naturals are positive and strictly increasing), converges to $0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences). Every term is defined and positive. This is the real-sequence counterpart of the rational null sequence $1 / n$ of the construction page, and it is the reference null sequence of this page: the later examples and counterexamples compute with it. The argument spends only density of $Q$ in $R$ (The rationals embed densely in the reals), the Archimedean property of $Q$ (The rationals are Archimedean) and reciprocal reversal (Inverses of positives are positive, and reciprocation reverses order); no completeness beyond the Archimedean property and no choice principle is used.

Facts & Assumptions

Given: A real number $ε > 0$ , and for every natural $k$ the real $x_{k} : = (k + 1)^{- 1}$ , where $k + 1 = σ (k)$ is the successor of $k$ read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

Density of $Q$ in $R$ : strictly between any two reals lies a rational, so for every real $ε > 0$ there is a rational $q$ with $0 < q < ε$ (The rationals embed densely in the reals).

[F2]

The rationals are Archimedean: for every rational $ε > 0$ there is a natural $n \geq 1$ with $n^{- 1} < ε$ (The rationals are Archimedean).

[F3]

Reciprocation reverses order on positives: $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ , and $a > 0$ implies $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[F4]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and the map $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there (Canonical naturals are positive and strictly increasing).

[F5]

Order is compatible with addition in $N$ : $m \leq n ⟺ m + k \leq n + k$ , and $m < n ⟺ m + k < n + k$ ; and $m \leq n$ with $m \neq n$ reads $m < n$ (Order is compatible with addition, Order on the natural numbers).

[F6]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Absolute value in an ordered field).

[F7]

Convergence: a sequence $(z_{k})$ of reals converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

Verification

technique · direct

1.1givenF1F2choose

Let a real $ε > 0$ be given. By [F1] choose a rational $q$ with $0 < q < ε$ , and by [F2] choose a natural $N \geq 1$ with $N^{- 1} < q$ .

2.1step 1.1F4F5algebra

For every natural $k \geq N$ one has $N + 1 \leq k + 1$ by [F5]. Read in $R$ through the canonical-natural embedding, [F4] gives $0 < N < N + 1$ since $N \geq 1$ , and $N + 1 \leq k + 1$ because $N + 1 \leq k + 1$ is either an equality of naturals, when the images agree, or a strict inequality of naturals at least $2$ , when [F4] makes the images strictly increase.

3.1step 2.1step 1.1F3algebra

For such $k$ , apply [F3] to $0 < N < N + 1$ to get $(N + 1)^{- 1} < N^{- 1}$ ; and $(k + 1)^{- 1} \leq (N + 1)^{- 1}$ because $N + 1 \leq k + 1$ is an equality or a strict inequality of positive reals, the strict case being [F3] again. Chaining with step 1.1, $(k + 1)^{- 1} \leq (N + 1)^{- 1} < N^{- 1} < q < ε$ .

4.1step 3.1F3F4F6

Every term is positive: $k + 1 \geq 1$ in $N$ , so $k + 1 > 0$ by [F4] and $(k + 1)^{- 1} > 0$ by [F3]; and $∣ x_{k} - 0 ∣ = ∣ x_{k} ∣ = (k + 1)^{- 1}$ by [F6]. With step 3.1 this gives $∣ x_{k} - 0 ∣ < ε$ for all $k \geq N$ .

5.1step 1.1step 4.1F7∎

Steps 1.1 through 4.1 were run for an arbitrary real $ε > 0$ , hence in particular for every rational $ε > 0$ , and the threshold $N$ produced witnesses the convergence condition of [F7]. Therefore $x_{k} \to 0$ .

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A rational expression in $k$ : $(2 (k + 1) + 3) / (5 (k + 1) + 1) \to 2 / 5$

Example

The sequence $x_{k} : = \frac{2 (k + 1) + 3}{5 (k + 1) + 1}$ converges to $2 / 5$ (Limits and Cauchy sequences of reals). The point is method: an explicit rational expression of $k$ is evaluated from the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients) once a single null sequence is known, without an $ε$ -estimate fitted to the expression. Dividing numerator and denominator by the positive factor $k + 1$ turns the expression into a quotient of two affine expressions in the null sequence $(k + 1)^{- 1}$ of The sequence $1 / (k + 1)$ converges to $0$ .

Facts & Assumptions

Given: The sequence $x_{k} : = (2 (k + 1) + 3) / (5 (k + 1) + 1)$ for $k \in N$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

The sequence $u_{k} : = (k + 1)^{- 1}$ converges to $0$ and every term $u_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Linearity clauses of the algebra of limits: if $(p_{k})$ converges to $p$ and $(q_{k})$ converges to $q$ , then $p_{k} + q_{k}$ converges to $p + q$ , and for every real $c$ the scalar multiple $c p_{k}$ converges to $c p$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F3]

Quotient clause of the algebra of limits: if $(p_{k})$ converges to $p$ and $(q_{k})$ converges to $q$ with $q \neq 0$ and $q_{k} \neq 0$ for every $k$ , then $p_{k} / q_{k}$ converges to $p / q$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F4]

Canonical naturals: $k + 1 \geq 1$ for every $k$ , and $n \mapsto n \cdot 1_{R}$ is strictly increasing and positive on ${1, 2, 3, \dots}$ , hence preserves non-strict order there; in particular $2, 3, 5$ denote positive reals (Canonical naturals are positive and strictly increasing).

[F5]

Field arithmetic: every nonzero element has a multiplicative inverse $u u^{- 1} = 1$ , multiplication distributes over addition, and $0 \neq 1$ (Field).

[F6]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ , so $∣ 0 ∣ = 0$ and $∣ c - c ∣ = ∣ 0 ∣ = 0$ for every real $c$ (Absolute value in an ordered field).

[F7]

Convergence: a sequence $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F9]

Order arithmetic: the positives are closed under addition and trichotomy holds, so an element that is a sum of positives is positive and in particular nonzero (Ordered field).

[F10]

$0 < 1$ in $R$ (The multiplicative identity is positive).

Verification

technique · direct

1.1givenF4F5F9algebra

For every $k$ the factor $k + 1$ is positive by [F4] and hence nonzero by trichotomy [F9], so multiplying numerator and denominator by $(k + 1)^{- 1}$ and distributing is legitimate: $x_{k} = (2 + 3 (k + 1)^{- 1}) / (5 + (k + 1)^{- 1})$ .

1.2givenF1

The sequence $u_{k} = (k + 1)^{- 1}$ converges to $0$ and every $u_{k}$ is positive.

1.3F6F7

For every real $c$ the constant sequence $c$ converges to $c$ : given a rational $ε > 0$ , the estimate $∣ c - c ∣ = ∣ 0 ∣ = 0 < ε$ of [F6] holds at every index, so the threshold $K = 0$ witnesses the condition of [F7].

2.1step 1.2step 1.3F2

The numerator converges: $3 u_{k} \to 3 \cdot 0 = 0$ by the scalar clause of [F2], and $2 + 3 u_{k} \to 2 + 0 = 2$ by the sum clause applied to the constant sequence $2$ of step 1.3 and $3 u_{k}$ .

3.1step 1.2step 1.3F2F4F9F10

The denominator converges and is nowhere zero: as in step 2.1, $5 + u_{k} \to 5 + 0 = 5$ by [F2] applied to the constant sequence $5$ of step 1.3 and $u_{k}$ ; and $5 + u_{k} > 0$ for every $k$ , since $5 > 0$ by [F4] and [F10], $u_{k} > 0$ by step 1.2, and sums of positives are positive by [F9].

4.1step 1.1step 2.1step 3.1F3F4F9F10∎

The limit of the denominator is nonzero: $5 > 0$ by [F4] and [F10], so $5 \neq 0$ by trichotomy [F9]. The quotient clause [F3] applied to the numerator of step 2.1 ( $\to 2$ ), the denominator of step 3.1 ( $\to 5$ , nowhere zero) and the value $5 \neq 0$ gives $x_{k} \to 2 / 5$ , using the identified form of step 1.1.

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The alternating sequence $(- 1)^{k} / (k + 1)$ converges to $0$

Example

Let $(s_{k})$ be the sequence with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ — the sequence usually written $(- 1)^{k}$ , introduced and proved non-convergent as a bounded sequence in FALSE: every bounded sequence converges. The sequence $y_{k} : = s_{k} (k + 1)^{- 1}$ converges to $0$ (Limits and Cauchy sequences of reals): a null sequence may oscillate in sign at every term. The proof is the null-times-bounded lemma (A null sequence times a bounded sequence is null) applied to the null sequence of The sequence $1 / (k + 1)$ converges to $0$ and the bounded sequence $(s_{k})$ , so no estimate depending on the sign pattern is needed.

Facts & Assumptions

Given: The unique sequence $(s_{k})$ of reals with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ (The recursion theorem), the sequence $y_{k} : = s_{k} (k + 1)^{- 1}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and the sequence $u_{k} : = (k + 1)^{- 1}$ .

[F1]

The recursion: $(s_{k})$ is the unique sequence with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ for every $k$ (FALSE: every bounded sequence converges).

[F2]

Induction principle: if $P (0)$ holds and $P (k)$ implies $P (σ (k))$ for every $k$ , then $P (k)$ holds for every natural $k$ (The principle of mathematical induction).

[F3]

Absolute value is even: $∣ - u ∣ = ∣ u ∣$ (Basic properties of the absolute value).

[F4]

Absolute value: $∣ u ∣ = u$ if $u \geq 0$ and $∣ u ∣ = - u$ if $u < 0$ (Absolute value in an ordered field).

[F5]

$(u_{k}) = ((k + 1)^{- 1})$ converges to $0$ and every term is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F6]

Null times bounded: if $(z_{k})$ converges to $0$ and $(w_{k})$ is bounded, then $(z_{k} w_{k})$ converges to $0$ (A null sequence times a bounded sequence is null).

[F7]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F8]

$0 < 1$ in $R$ (The multiplicative identity is positive).

Verification

technique · direct

1.1givenF1F2F3F4F8

By induction [F2], $∣ s_{k} ∣ = 1$ for every $k$ : the base is $∣ s_{0} ∣ = ∣ 1 ∣ = 1$ , valid because $1 \geq 0$ by [F8] and $∣ u ∣ = u$ for $u \geq 0$ by [F4]; and if $∣ s_{k} ∣ = 1$ then $∣ s_{σ (k)} ∣ = ∣ - s_{k} ∣ = ∣ s_{k} ∣ = 1$ by the recursion [F1] and [F3].

1.2givenF5

The sequence $(u_{k}) = ((k + 1)^{- 1})$ converges to $0$ .

2.1step 1.1F7

Hence $(s_{k})$ is bounded with bound $1$ : $∣ s_{k} ∣ = 1 \leq 1$ for every $k$ , which is the condition of [F7] with $M = 1$ .

3.1step 2.1step 1.2F6∎

The product $y_{k} = s_{k} u_{k}$ is the product of the bounded sequence $(s_{k})$ of step 2.1 and the null sequence $(u_{k})$ of step 1.2, so [F6] gives $y_{k} \to 0$ ; no hypothesis on the sign pattern of $(s_{k})$ beyond boundedness is used.

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The square roots $\sqrt{k}$ diverge to $+ \infty$

Example

Let $\sqrt{k}$ denote the unique nonnegative square root of the canonical natural $k \in R$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Then $(\sqrt{k})_{k \in N}$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ): it is unbounded, hence not convergent (Every convergent sequence is bounded). The proof is the Archimedean comparison $k > M^{2} \Rightarrow \sqrt{k} > M$ ; it uses only that $R$ is Archimedean (Every complete ordered field is Archimedean) and that squaring is monotone on the nonnegatives (Squaring is monotone on the nonnegatives).

Facts & Assumptions

Given: For every natural $k$ the canonical natural $k \cdot 1_{R}$ , written $k$ , and the real $\sqrt{k}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

Divergence to $+ \infty$ : $z_{k} \to + \infty$ means that for every real $M$ there is $K$ with $z_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F2]

Archimedean property: for every real $x$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F3]

Canonical naturals: $0 \cdot 1_{R} = 0$ and $n \cdot 1_{R} > 0$ for $n \geq 1$ ; the map $n \mapsto n \cdot 1_{R}$ is strictly increasing, hence preserves non-strict order, on ${1, 2, 3, \dots}$ ; so the real $k = k \cdot 1_{R}$ is nonnegative for every natural $k$ (The unique embedding of ℚ into an ordered field, Canonical naturals are positive and strictly increasing).

[F4]

Square roots: every real $a \geq 0$ has a unique $s \geq 0$ with $s^{2} = a$ , written $s = \sqrt{a}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F5]

Monotonicity of squaring on nonnegatives: for $a, b \geq 0$ , $a < b ⟺ a^{2} < b^{2}$ (Squaring is monotone on the nonnegatives).

[F6]

Trichotomy and order arithmetic: for all reals $u, v$ exactly one of $u < v$ , $u = v$ , $v < u$ holds, and $u \geq 0 > v$ implies $u > v$ (Ordered field, Order on the reals).

[F7]

A sequence converging to a real is bounded, so a sequence that diverges to $+ \infty$ — being unbounded, since it eventually exceeds every real bound — is not convergent (Every convergent sequence is bounded, Divergence to $+ \infty$ and to $- \infty$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Verification

technique · direct

1.1givenF3F4

For every natural $k$ the real $k$ is nonnegative by [F3], so [F4] applies: $\sqrt{k}$ is defined with $\sqrt{k} \geq 0$ and $(\sqrt{k})^{2} = k$ .

2.1step 1.1F1F6

First case, $M < 0$ : by step 1.1, $\sqrt{k} \geq 0 > M$ for every $k$ , so $\sqrt{k} > M$ for all $k$ and the threshold $K = 0$ witnesses the condition of [F1] for this $M$ .

2.2step 1.1F2F3F5F6

Second case, $M \geq 0$ : apply [F2] to the real $M^{2}$ to obtain a natural $n \geq 1$ with $M^{2} < n$ . For every $k \geq n$ the order preservation of [F3] gives $n \leq k$ , hence $M^{2} < k$ ; since $M \geq 0$ and $\sqrt{k} \geq 0$ with $(\sqrt{k})^{2} = k$ by step 1.1, the equivalence of [F5] at the nonnegative pair $M, \sqrt{k}$ gives $M < \sqrt{k}$ ; so every $k \geq n$ satisfies $\sqrt{k} > M$ .

3.1step 2.1step 2.2F1F6F7∎

Steps 2.1 and 2.2 dispose of the two cases $M < 0$ and $M \geq 0$ , which exhaust the reals by [F6]: for every real $M$ there is a threshold beyond which $\sqrt{k} > M$ , which is exactly the condition of [F1]; so $(\sqrt{k})$ diverges to $+ \infty$ . By [F7] it is therefore unbounded and does not converge.

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The successive differences $\sqrt{k + 1} - \sqrt{k}$ converge to $0$

Example

The sequence $z_{k} : = \sqrt{k + 1} - \sqrt{k}$ converges to $0$ (Limits and Cauchy sequences of reals). Since $(\sqrt{k + 1} - \sqrt{k}) (\sqrt{k + 1} + \sqrt{k}) = 1$ , the differences are the reciprocals of the sequence $w_{k} : = \sqrt{k + 1} + \sqrt{k}$ , which diverges to $+ \infty$ because $w_{k} \geq \sqrt{k}$ and $(\sqrt{k})$ does (The square roots $\sqrt{k}$ diverge to $+ \infty$ ); the backward direction of For positive terms, null and divergence to $+ \infty$ are reciprocal then gives $z_{k} \to 0$ . Thus a difference of two sequences that both diverge can converge, so the algebra of limits has no difference rule for divergent sequences.

Facts & Assumptions

Given: For every natural $k$ the reals $\sqrt{k}$ and $\sqrt{k + 1}$ , with $k + 1 = σ (k)$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and the sequences $z_{k} : = \sqrt{k + 1} - \sqrt{k}$ , $w_{k} : = \sqrt{k + 1} + \sqrt{k}$ .

[F1]

$(\sqrt{k})_{k \in N}$ diverges to $+ \infty$ (The square roots $\sqrt{k}$ diverge to $+ \infty$ ).

[F2]

Reciprocal bridge: for a sequence $(x_{k})$ with $x_{k} > 0$ for every $k$ , $(x_{k})$ converges to $0$ if and only if $(1 / x_{k})$ diverges to $+ \infty$ (For positive terms, null and divergence to $+ \infty$ are reciprocal).

[F3]

[F4]

Divergence to $+ \infty$ : $x_{k} \to + \infty$ means that for every real $M$ there is $K$ with $x_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F5]

Field arithmetic: every nonzero element has a multiplicative inverse, and $u v = 1$ with $v$ the inverse of $u$ identifies $v = u^{- 1}$ (Field).

[F6]

Monotonicity of squaring on nonnegatives: for $a, b \geq 0$ , $a < b ⟺ a^{2} < b^{2}$ and $a \leq b ⟺ a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

[F7]

Adding inequalities: if $a < b$ and $c < d$ then $a + c < b + d$ , and translation by a constant preserves the order (Order is preserved by adding a constant and by adding inequalities).

[F8]

$0 < 1$ (The multiplicative identity is positive).

[F9]

Archimedean property: for every real $x$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F10]

Canonical naturals: $k + 1 \geq 1$ for every $k$ , and $n \mapsto n \cdot 1_{R}$ is strictly increasing, hence preserves non-strict order, on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[F11]

Inverses of positives are positive: $a > 0$ implies $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

Verification

technique · direct

1.1givenF3F5F6F7F8F10

For every $k$ , $(\sqrt{k + 1})^{2} = k + 1$ and $(\sqrt{k})^{2} = k$ by [F3], so the product of the two factors is $z_{k} w_{k} = (\sqrt{k + 1} - \sqrt{k}) (\sqrt{k + 1} + \sqrt{k}) = (k + 1) - k = 1$ by [F5] and algebra. Moreover $w_{k} > 0$ : both roots are nonnegative by [F3], $1 \leq k + 1$ by [F10] with $1^{2} = 1 \leq k + 1 = (\sqrt{k + 1})^{2}$ gives $1 \leq \sqrt{k + 1}$ by [F6], and adding the nonnegative $\sqrt{k}$ to it preserves the inequality by [F7], so $w_{k} \geq \sqrt{k + 1} \geq 1 > 0$ by [F8]. Since $w_{k} \neq 0$ , the identity $z_{k} w_{k} = 1$ exhibits $z_{k}$ as the inverse of $w_{k}$ by [F5].

1.2givenF3F7

For every $k$ , $0 \leq \sqrt{k + 1}$ by [F3]; adding $\sqrt{k}$ to this inequality gives $\sqrt{k} \leq \sqrt{k + 1} + \sqrt{k} = w_{k}$ . Translation preserves a non-strict inequality by [F7] in the strict case and by equality otherwise.

1.3givenF3F4F6F9F10

The shifted sequence $(\sqrt{k + 1})$ diverges to $+ \infty$ : given a real $M$ , if $M < 0$ then $\sqrt{k + 1} \geq 0 > M$ for every $k$ by [F3]; if $M \geq 0$ , [F9] gives a natural $n \geq 1$ with $M^{2} < n$ , and for every $k \geq n$ one has $n \leq k + 1$ by [F10], hence $M^{2} < (\sqrt{k + 1})^{2}$ ; since $M \geq 0$ and $\sqrt{k + 1} \geq 0$ , the equivalence of [F6] gives $M < \sqrt{k + 1}$ . So for every real $M$ the bound holds eventually, which is [F4].

2.1step 1.2F1F4

The sequence $(w_{k})$ diverges to $+ \infty$ : by [F1] the sequence $(\sqrt{k})$ does, so given a real $M$ there is $K$ with $\sqrt{k} > M$ for all $k \geq K$ , and step 1.2 gives $w_{k} \geq \sqrt{k} > M$ there; this is the condition of [F4].

3.1step 1.1step 1.3step 2.1F1F2F5F11∎

Put $x_{k} : = 1 / w_{k}$ . Since $w_{k} > 0$ for every $k$ by step 1.1, [F11] gives $x_{k} > 0$ for every $k$ , and $1 / x_{k} = 1 / (1 / w_{k}) = w_{k}$ by [F5]. Now $1 / x_{k} = w_{k}$ diverges to $+ \infty$ by step 2.1, so the backward direction of the reciprocal bridge [F2], applied to the positive sequence $(x_{k})$ , gives $x_{k} = 1 / w_{k} \to 0$ ; and $z_{k} = w_{k}^{- 1} = 1 / w_{k}$ by step 1.1, so $z_{k} \to 0$ . Moreover $(\sqrt{k})$ diverges to $+ \infty$ by [F1] and $(\sqrt{k + 1})$ does by step 1.3, so $z_{k}$ is the difference of two sequences that both diverge to $+ \infty$ and yet converges to $0$ : the algebra of limits has no difference rule for divergent sequences.

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Changing the first hundred terms leaves convergence to $0$ and boundedness intact

Example

Let $u_{k} : = (k + 1)^{- 1}$ and let $(x_{k})$ be $u$ modified in its first hundred terms: $x_{k} : = k + 1$ for $k < 100$ and $x_{k} : = (k + 1)^{- 1}$ for $k \geq 100$ . Then $x_{k} \to 0$ and $(x_{k})$ is bounded, indeed $∣ x_{k} ∣ \leq 100$ for every $k$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). The initial block is arbitrary prescribed data: convergence depends only on the tail (Convergence depends only on the tail), while the bound is read off the finite block and the tail separately. This is the worked form of the tail theorem.

Facts & Assumptions

Given: The sequence $u_{k} : = (k + 1)^{- 1}$ of The sequence $1 / (k + 1)$ converges to $0$ and the sequence $x_{k} : = k + 1$ for $k < 100$ , $x_{k} : = (k + 1)^{- 1}$ for $k \geq 100$ , where $k + 1 = σ (k)$ and $100$ are read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F2]

Tails and convergence: $(x_{k})$ converges to $x$ if and only if every tail $x^{(K)}$ converges to $x$ ; consequently, if two sequences agree from some index on, then either both converge to $x$ or neither does (Convergence depends only on the tail).

[F3]

The reference null sequence: $u_{k} = (k + 1)^{- 1}$ converges to $0$ , with every term positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F4]

Discreteness: for all naturals $m, n$ , $m < n ⟺ σ (m) \leq n$ , so there is no natural strictly between $n$ and $σ (n)$ (Discreteness: $σ (n)$ is the immediate successor, Order on the natural numbers).

[F5]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there; in particular $1 \leq 100$ and every $k + 1 \geq 1$ maps to a positive real (Canonical naturals are positive and strictly increasing).

[F6]

Reciprocals against $1$ : for $r > 0$ one has $r \geq 1 ⟺ 0 < r^{- 1} \leq 1$ , and $r = 1 ⟺ r^{- 1} = 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[F7]

Inverses of positives are positive: $a > 0$ implies $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[F8]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Absolute value in an ordered field).

[F9]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F10]

The order on $N$ is total: for all naturals $m, n$ exactly one of $m < n$ , $m = n$ , $n < m$ holds ( $\leq$ is a linear order on $N$ ).

Verification

technique · direct

1.1given

For every $k \geq 100$ the second clause of the definition of $(x_{k})$ gives $x_{k} = (k + 1)^{- 1} = u_{k}$ , so $(x_{k})$ and $(u_{k})$ agree from index $100$ on.

2.1step 1.1F2F3

By [F3] the sequence $(u_{k})$ converges to $0$ ; by the agreement clause of [F2] the sequence $(x_{k})$ converges to $0$ as well.

2.2step 1.1F4F5F8

For $k < 100$ : discreteness [F4] gives $k + 1 = σ (k) \leq 100$ , and $k + 1 \geq 1$ , so the real $x_{k} = k + 1$ satisfies $1 \leq x_{k} \leq 100$ by [F5] and is positive; hence $∣ x_{k} ∣ = x_{k} \leq 100$ by [F8].

2.3step 1.1F5F6F7F8

For $k \geq 100$ : the real $k + 1$ satisfies $k + 1 \geq 100 \geq 1$ , so it is positive with $(k + 1)^{- 1} \leq 1$ by [F6] and $(k + 1)^{- 1} > 0$ by [F7]; since $1 \leq 100$ by [F5], transitivity gives $0 < x_{k} \leq 100$ , hence $∣ x_{k} ∣ = x_{k} \leq 100$ by [F8].

3.1step 2.1step 2.2step 2.3F9F10∎

Every natural $k$ satisfies $k < 100$ or $k \geq 100$ by totality [F10], and steps 2.2 and 2.3 give $∣ x_{k} ∣ \leq 100$ in the two cases; so $100$ is a bound in the sense of [F9] and $(x_{k})$ is bounded. With step 2.1 this shows that changing the first hundred terms leaves both convergence to $0$ and boundedness, with the explicit bound $100$ .

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Two squeeze computations: $1 / ((k + 1) \sqrt{k + 1}) \to 0$ and $(k + (- 1)^{k}) / (k + 1) \to 1$

Example

(i) The sequence $a_{k} : = ((k + 1) \sqrt{k + 1})^{- 1}$ converges to $0$ , squeezed between the constant sequence $0$ and $(k + 1)^{- 1}$ . (ii) With $(s_{k})$ the alternating sequence of FALSE: every bounded sequence converges, the sequence $b_{k} : = (k + s_{k}) / (k + 1)$ converges to $1$ , squeezed between $1 - 2 (k + 1)^{- 1}$ and the constant $1$ . Both use The squeeze theorem; in (ii) the middle sequence is not presented as a sum of two sequences whose limits are known in advance, which is what the squeeze theorem is for.

Facts & Assumptions

Given: For every natural $k$ the reals $\sqrt{k + 1}$ with $k + 1 = σ (k)$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the sequence $u_{k} : = (k + 1)^{- 1}$ , and the alternating sequence $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ (FALSE: every bounded sequence converges); the sequences $a_{k} : = ((k + 1) \sqrt{k + 1})^{- 1}$ and $b_{k} : = (k + s_{k}) / (k + 1)$ .

[F1]

Squeeze theorem: if $a_{k} \leq x_{k} \leq b_{k}$ eventually and $(a_{k})$ , $(b_{k})$ both converge to the same real $L$ , then $(x_{k})$ converges to $L$ (The squeeze theorem).

[F2]

Sum and scalar clauses of the algebra of limits: if $(p_{k})$ converges to $p$ and $(q_{k})$ converges to $q$ , then $p_{k} + q_{k}$ converges to $p + q$ ; and $c p_{k}$ converges to $c p$ for every real $c$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F3]

$(u_{k}) = ((k + 1)^{- 1})$ converges to $0$ and every $u_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F4]

The recursion $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ defines the alternating sequence $(s_{k})$ (FALSE: every bounded sequence converges).

[F5]

Induction principle (The principle of mathematical induction).

[F6]

Square roots: every real $a \geq 0$ has a unique $s \geq 0$ with $s^{2} = a$ , written $\sqrt{a}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F7]

Monotonicity of squaring on nonnegatives: for $a, b \geq 0$ , $a \leq b ⟺ a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

[F8]

Inverses of positives: $a > 0$ implies $a^{- 1} > 0$ ; and $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order).

[F9]

Multiplying inequalities of nonnegatives: $0 \leq x \leq y$ and $0 \leq u \leq v$ imply $x u \leq y v$ (Multiplying inequalities of positives).

[F10]

Sign rules: a product of two nonnegatives is nonnegative, and for $c > 0$ , $x < y$ implies $x c < y c$ (Sign rules for products and monotonicity of multiplication).

[F11]

Field arithmetic: every nonzero element has an inverse and multiplication distributes over addition (Field).

[F12]

Order arithmetic: the positives are closed under addition and trichotomy holds, so $2 : = 1 + 1$ is positive and $2 \neq 0$ (Ordered field, The multiplicative identity is positive).

[F13]

Convergence: $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F14]

Canonical naturals: $k + 1 \geq 1$ and $k + 1 > 0$ , hence $k + 1 \neq 0$ , for every natural $k$ ; the map $n \mapsto n \cdot 1_{R}$ is strictly increasing, hence preserves non-strict order, on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[F15]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ , so $∣ 0 ∣ = 0$ (Absolute value in an ordered field).

Verification

technique · direct

1.1F13F15

For every real $c$ the constant sequence $c$ converges to $c$ : given a rational $ε > 0$ , $∣ c - c ∣ = ∣ 0 ∣ = 0 < ε$ at every index by [F15], so the threshold $K = 0$ witnesses [F13].

1.2givenF3

The sequence $(u_{k}) = ((k + 1)^{- 1})$ converges to $0$ with every term positive.

1.3givenF4F5

By induction [F5], $s_{k} \in {1, - 1}$ for every $k$ : $s_{0} = 1$ ; and if $s_{k} = \pm 1$ then $s_{σ (k)} = - s_{k} = \mp 1$ by [F4].

1.4givenF6F7F14

For every $k$ one has $\sqrt{k + 1} \geq 1$ : by [F6], $\sqrt{k + 1} \geq 0$ and $(\sqrt{k + 1})^{2} = k + 1$ , while $1 \leq k + 1$ by [F14] with $1^{2} = 1 \leq k + 1$ , so the equivalence [F7] at the nonnegative pair $1, \sqrt{k + 1}$ gives $1 \leq \sqrt{k + 1}$ .

2.1step 1.4F8F9F10F14

For every $k$ , $0 \leq a_{k} \leq u_{k}$ : put $A : = (k + 1) \sqrt{k + 1}$ and $B : = k + 1$ , so that $A - B = (k + 1) (\sqrt{k + 1} - 1) \geq 0$ by [F9] and [F10], because both factors are nonnegative by [F14] and step 1.4; hence $0 < B \leq A$ , and reciprocal reversal [F8] gives $a_{k} = A^{- 1} \leq B^{- 1} = u_{k}$ , while $a_{k} > 0$ by [F8].

2.2step 1.3givenF3F10F11F12F14

For every $k$ , the denominator $k + 1$ is nonzero by [F14], and the middle sequence is bounded by the two claimed bounds: if $s_{k} = 1$ then $b_{k} = (k + 1) / (k + 1) = 1$ ; if $s_{k} = - 1$ then $b_{k} = (k - 1) / (k + 1) = 1 - 2 (k + 1)^{- 1} = 1 - 2 u_{k}$ . In both cases $1 - 2 u_{k} \leq b_{k} \leq 1$ : since $2 u_{k} > 0$ by [F3], [F10] and [F12], one has $1 - 2 u_{k} < 1$ . If $s_{k} = 1$ the upper bound is attained and the lower inequality is strict; if $s_{k} = - 1$ the lower bound is attained and the upper inequality is strict.

2.3step 1.1step 1.2F2

The lower bound converges: $- 2 u_{k} \to (- 2) \cdot 0 = 0$ by the scalar clause of [F2] applied to step 1.2, and $1 - 2 u_{k} = 1 + (- 2 u_{k}) \to 1 + 0 = 1$ by the sum clause applied to the constant sequence $1$ of step 1.1 and $(- 2 u_{k})$ .

3.1step 1.1step 1.2step 2.1F1

Squeeze for $a_{k}$ : the constant sequence $0$ converges to $0$ by step 1.1, $(u_{k})$ converges to $0$ by step 1.2, and $0 \leq a_{k} \leq u_{k}$ at every index by step 2.1; so [F1] gives $a_{k} \to 0$ .

4.1step 1.1step 2.2step 2.3F1∎

Squeeze for $b_{k}$ : the lower bound converges to $1$ by step 2.3, the constant sequence $1$ converges to $1$ by step 1.1, and $1 - 2 u_{k} \leq b_{k} \leq 1$ at every index by step 2.2; so [F1] gives $b_{k} \to 1$ .

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Even and odd subsequences of $1 / (k + 1)$ converge to $0$

Example

Let $(x_{k})$ be the null sequence of The sequence $1 / (k + 1)$ converges to $0$ , let $n : N \to N$ be the even strictly increasing index map constructed in FALSE: every bounded sequence converges, and let $m_{j} : = σ (n_{j})$ be its successor, the odd index map. Then the subsequences $x_{n_{j}}$ and $x_{m_{j}}$ both converge to $0$ (Subsequences inherit the limit): removing infinitely many indices from this already convergent sequence, even all odd ones, preserves its inherited limit. A convergent subsequence of a divergent sequence can exist, so no converse is asserted, and by A strictly increasing index map satisfies $n_{k} \geq k$ the index maps satisfy $n_{j} \geq j$ and $m_{j} \geq j$ , so the subsequences are genuine thinnings of the sequence.

Facts & Assumptions

Given: The null sequence $x_{k} : = (k + 1)^{- 1}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the index map $n$ with $n_{0} = 0$ and $n_{σ (j)} = σ (σ (n_{j}))$ constructed in FALSE: every bounded sequence converges, and the successor map $m_{j} : = σ (n_{j})$ .

[F1]

$(x_{k})$ converges to $0$ (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

The map $n$ satisfies $n_{0} = 0$ , $n_{σ (j)} = σ (σ (n_{j}))$ for every $j$ , and $n$ is strictly increasing (FALSE: every bounded sequence converges).

[F3]

Subsequences inherit limits: if $(z_{k})$ converges to $z$ and $p : N \to N$ is strictly increasing, then $(z_{p_{j}})$ converges to $z$ (Subsequences inherit the limit).

[F4]

Growth of index maps: if $p$ is strictly increasing then $p_{j} \geq j$ for every $j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[F5]

Consecutive comparisons suffice for strict increase: if $p_{i} < p_{σ (i)}$ for every $i$ , then $p$ is strictly increasing (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[F6]

Strictly increasing means $p_{i} < p_{j}$ whenever $i < j$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F7]

Discreteness: for naturals $a, b$ , $a < b ⟺ σ (a) \leq b$ (Discreteness: $σ (n)$ is the immediate successor).

Verification

technique · direct

1.1givenF1F2

$(x_{k})$ converges to $0$ and $n$ is strictly increasing.

1.2givenF2F5F7

For every $j$ one has $m_{j} = σ (n_{j}) = n_{j} + 1$ , so $m$ is strictly increasing: from $n_{j} < n_{σ (j)}$ and [F7] we get $m_{j} = σ (n_{j}) \leq n_{σ (j)} < σ (n_{σ (j)}) = m_{σ (j)}$ , that is $m_{j} < m_{σ (j)}$ at every $j$ , so [F5] applies.

2.1step 1.1F3

The subsequence along $n$ converges to $0$ : [F3] applied to step 1.1 gives $x_{n_{j}} \to 0$ .

2.2step 1.1step 1.2F3

The subsequence along $m$ converges to $0$ : [F3] applied to step 1.1 and the strict increase of step 1.2 gives $x_{m_{j}} \to 0$ .

3.1step 1.1step 1.2step 2.1step 2.2F4F6∎

Both index maps grow at least like the identity: $n_{j} \geq j$ by [F4] applied to the strict increase of $n$ in step 1.1, and $m_{j} \geq j$ by [F4] applied to the strict increase of $m$ in step 1.2. Hence the even and odd subsequences of [F6] are genuine thinnings of $(x_{k})$ and both converge to the inherited limit $0$ .

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The absolute value of the limit, and the limit of the absolute value

Example

Let $x_{k} : = (k + 1)^{- 1} - 1$ . Then $x_{k} \to - 1$ , and $∣ x_{k} ∣ \to 1 = ∣ - 1 ∣$ (The absolute value is compatible with limits, Limits and Cauchy sequences of reals). Since $∣ x_{k} ∣ = 1 - (k + 1)^{- 1}$ converges to $1$ while $x_{k}$ converges to $- 1 \neq 1$ , the magnitude of the limit is recovered from the absolute values, but its sign is not. In particular, $∣ x_{k} ∣ \to ∣ 1 ∣$ does not imply $x_{k} \to 1$ : the implication of The absolute value is compatible with limits cannot be reversed for the candidate limit $1$ .

Facts & Assumptions

Given: The sequences $x_{k} : = (k + 1)^{- 1} - 1$ and $u_{k} : = (k + 1)^{- 1}$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F1]

$(u_{k})$ converges to $0$ and every $u_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Sum clause of the algebra of limits: if $(p_{k})$ converges to $p$ and $(q_{k})$ converges to $q$ , then $p_{k} + q_{k}$ converges to $p + q$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F3]

Absolute values of limits: if $(z_{k})$ converges to $z$ then $(∣ z_{k} ∣)$ converges to $∣ z ∣$ (The absolute value is compatible with limits).

[F4]

Uniqueness of limits: if $(z_{k})$ converges to $z$ and to $z^{'}$ , then $z = z^{'}$ (A sequence has at most one limit).

[F5]

Absolute value: $∣ u ∣ = u$ if $u \geq 0$ and $∣ u ∣ = - u$ if $u < 0$ ; so $∣ 0 ∣ = 0$ and $∣ - 1 ∣ = - (- 1) = 1$ because $- 1 < 0$ (Absolute value in an ordered field).

[F6]

Reciprocals against $1$ : for $r > 0$ , $r \geq 1 ⟺ 0 < r^{- 1} \leq 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[F7]

$0 < 1$ in $R$ (The multiplicative identity is positive).

[F8]

Translation invariance: $a < b$ implies $a + c < b + c$ (Order is preserved by adding a constant and by adding inequalities).

[F9]

Convergence: $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F10]

Field arithmetic: $u - v = u + (- v)$ , $- (u - v) = v - u$ , and $- (- 1) = 1$ (Field).

[F11]

Trichotomy: for all reals $u, v$ exactly one of $u < v$ , $u = v$ , $v < u$ holds (Ordered field, Order on the reals).

Verification

technique · direct

1.1givenF1

The sequence $(u_{k})$ converges to $0$ with every term positive.

1.2F5F9

For every real $c$ the constant sequence $c$ converges to $c$ : given a rational $ε > 0$ and any index $k$ , $∣ c - c ∣ = ∣ 0 ∣ = 0 < ε$ by [F5], so the threshold $K = 0$ witnesses the condition of [F9].

2.1step 1.1step 1.2F2

The sum $x_{k} = u_{k} + (- 1)$ converges to $0 + (- 1) = - 1$ by the sum clause [F2] applied to step 1.1 and to the constant sequence $- 1$ of step 1.2.

2.2step 1.1F1F5F6F10F11

For every $k$ one has $0 < u_{k} \leq 1$ : positivity is [F1], and $u_{k} \leq 1$ is the case $r = k + 1 \geq 1$ of [F6]. Hence $x_{k} = u_{k} - 1 \leq 0$ . If $x_{k} < 0$ , the negative clause of [F5] gives $∣ x_{k} ∣ = - x_{k}$ ; if $x_{k} = 0$ , both sides are $0$ . Thus $∣ x_{k} ∣ = - x_{k} = 1 - u_{k}$ at every index, including $k = 0$ .

3.1step 2.1F3F5F7F8F10

Hence $∣ x_{k} ∣ \to ∣ - 1 ∣ = 1$ : step 2.1 and [F3] give $∣ x_{k} ∣ \to ∣ - 1 ∣$ , and $∣ - 1 ∣ = 1$ because $- 1 < 0$ by [F7] and [F8], so [F5] gives $∣ - 1 ∣ = - (- 1) = 1$ by [F10].

4.1step 2.1step 3.1F3F4F5F7F8F11∎

Consider the converse of [F3] for the candidate limit $c = 1$ : it would assert that $∣ x_{k} ∣ \to ∣ 1 ∣ = 1$ implies $x_{k} \to 1$ . Its hypothesis holds by step 3.1, but its conclusion would contradict $x_{k} \to - 1$ from step 2.1, since uniqueness [F4] would force $- 1 = 1$ , whereas $- 1 < 0 < 1$ by [F7] and [F8]. Thus this witness refutes the converse for the nonzero candidate $1$ , while satisfying the forward implication for its actual limit $- 1$ .

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The sum of two divergent sequences can converge: $k + (- k)$

Statement refuted

Refuted claim: if the sum $(x_{k} + y_{k})$ of two sequences of reals converges, then $(x_{k})$ and $(y_{k})$ both converge.

The witnesses are $x_{k} : = k$ and $y_{k} : = - k$ , where $k$ denotes the canonical natural of $R$ (Canonical naturals are positive and strictly increasing): $(x_{k})$ diverges to $+ \infty$ and $(y_{k})$ to $- \infty$ , while the sum is the constant sequence $0$ , which converges to $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). This is the caution noted immediately after the algebra of limits: the clauses of Algebra of limits: sums, scalar multiples, products and quotients may not be read backwards.

Facts & Assumptions

Given: The sequences $x_{k} : = k$ and $y_{k} : = - k$ , where the natural $k$ denotes its canonical image $k \cdot 1_{R}$ in $R$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

Divergence to signed infinity: $z_{k} \to + \infty$ when for every real $M$ there is $K$ with $z_{k} > M$ for all $k \geq K$ ; and $z_{k} \to - \infty$ when for every real $M$ there is $K$ with $z_{k} < M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F2]

Archimedean property: for every real $x$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F3]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there (Canonical naturals are positive and strictly increasing).

[F4]

Order arithmetic in an ordered field: $a < b$ means $b - a$ is positive, so $a < b$ if and only if $- b < - a$ ; and trichotomy holds for the strict order (Ordered field, Order on the reals).

[F5]

Field arithmetic: every element $u$ has an additive inverse $- u$ with $u + (- u) = 0$ , and $u - u = 0$ (Field).

[F6]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ , so $∣ 0 ∣ = 0$ (Absolute value in an ordered field).

[F7]

Convergence: a sequence $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F8]

Every convergent sequence is bounded: if $(z_{k})$ converges to $z$ then there is $M$ with $∣ z_{k} ∣ \leq M$ for every $k$ (Every convergent sequence is bounded).

[F9]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F10]

A sequence diverging to $+ \infty$ or to $- \infty$ is unbounded, and an unbounded sequence is not convergent — the latter being the contrapositive of [F8] (Divergence to $+ \infty$ and to $- \infty$ , Every convergent sequence is bounded, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F11]

Sum rule of the algebra of limits: if $(u_{k})$ converges to $u$ and $(v_{k})$ converges to $v$ , then $u_{k} + v_{k}$ converges to $u + v$ (Algebra of limits: sums, scalar multiples, products and quotients).

Counterexample

technique · direct

1.1givenF1F2F3

$(x_{k}) = (k)$ diverges to $+ \infty$ : given a real $M$ , [F2] supplies a natural $N \geq 1$ with $M < N$ , and for every $k \geq N$ the order preservation of [F3] gives $N \leq k$ , so $x_{k} > M$ ; this is the condition of [F1] with $K = N$ .

1.2givenF1F2F3F4

$(y_{k}) = (- k)$ diverges to $- \infty$ : given a real $M$ , apply [F2] to $- M$ to get a natural $N \geq 1$ with $- M < N$ ; for every $k \geq N$ order preservation [F3] gives $N \leq k$ , and [F4] turns $- M < k$ into $- k < M$ , so $y_{k} < M$ for all $k \geq N$ .

1.3givenF5

For every $k$ the sum is $x_{k} + y_{k} = k + (- k) = 0$ , so the sum sequence is constantly $0$ .

2.1step 1.3F6F7

The constant sequence $0$ converges to $0$ : for a rational $ε > 0$ and every $k$ , $∣ 0 - 0 ∣ = ∣ 0 ∣ = 0 < ε$ by [F6], so the threshold $K = 0$ witnesses the condition of [F7].

3.1step 1.1step 1.2step 2.1F8F9F10F11∎

Neither summand converges: by steps 1.1 and 1.2 the sequences diverge to signed infinity, so by [F10] they are unbounded in the sense of [F9] and do not converge. Meanwhile the sum converges to $0$ by step 2.1, although the convergence hypothesis of the sum rule [F11] fails for both summands; hence the sum rule cannot be read backwards and the refuted claim fails.

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A null sequence times an unbounded sequence need not be null: $(k + 1)^{- 1} \cdot (k + 1) = 1$

Statement refuted

Refuted claim: if $(x_{k})$ converges to $0$ , then the product $x_{k} y_{k}$ converges to $0$ for every sequence $(y_{k})$ of reals.

The witnesses are $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = k + 1$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Canonical naturals are positive and strictly increasing): $(x_{k})$ is null (The sequence $1 / (k + 1)$ converges to $0$ ) and $(y_{k})$ diverges to $+ \infty$ , but $x_{k} y_{k} = 1$ for every $k$ , so the product does not converge to $0$ : its distance from $0$ is $1$ at every index, and the rational test value $ε = 1$ never passes. The boundedness hypothesis of A null sequence times a bounded sequence is null therefore cannot be dropped; it is the only hypothesis that lemma uses beyond convergence of the null factor.

Facts & Assumptions

Given: The sequences $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = k + 1$ , where $k + 1 = σ (k)$ is the successor of $k$ read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

$(x_{k})$ converges to $0$ and every term $x_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Divergence to $+ \infty$ : $y_{k} \to + \infty$ means that for every real $M$ there is $K$ with $y_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F3]

Archimedean property: for every real $x$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F4]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there (Canonical naturals are positive and strictly increasing).

[F5]

Order is compatible with addition in $N$ : for all naturals $m, n, k$ one has $m \leq n ⟺ m + k \leq n + k$ (Order is compatible with addition).

[F6]

Field arithmetic: every nonzero element $u$ has a multiplicative inverse $u^{- 1}$ with $u \cdot u^{- 1} = 1$ (Field).

[F7]

The boundedness hypothesis: if $(x_{k})$ converges to $0$ and $(y_{k})$ is bounded, then $(x_{k} y_{k})$ converges to $0$ (A null sequence times a bounded sequence is null).

[F8]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Absolute value in an ordered field).

[F9]

Convergence: a sequence $(z_{k})$ of reals converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F10]

Strict order: for all reals $u, v$ exactly one of $u < v$ , $u = v$ , $v < u$ holds, so $u < u$ is false (Ordered field).

[F11]

$0 < 1$ in $R$ (The multiplicative identity is positive).

[F12]

The map $q \mapsto \hat{q}$ is an embedding of ordered fields, so the rational $1$ is identified with the real $1$ (The rationals embed densely in the reals).

[F13]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Counterexample

technique · direct

1.1givenF2F3F4F5

$(y_{k}) = (k + 1)$ diverges to $+ \infty$ : given a real $M$ , [F3] supplies a natural $N \geq 1$ with $M < N$ , and for every $k \geq N$ one has $N + 1 \leq k + 1$ by [F5], hence $N < N + 1 \leq k + 1$ by [F4], so $y_{k} = k + 1 > M$ ; this is exactly the condition of [F2] with $K = N$ .

1.2givenF1

$(x_{k})$ converges to $0$ and every term $x_{k}$ is positive by [F1].

1.3givenF4F6algebra

For every $k$ the factor $k + 1$ is nonzero, and field arithmetic [F6] gives $x_{k} y_{k} = (k + 1)^{- 1} (k + 1) = 1$ .

2.1step 1.3F8F9F10F11F12algebra

The product sequence is therefore constantly $1$ . At the rational test value $ε : = 1$ , identified with its real image by [F12] and positive by [F11], the convergence condition of [F9] would require $∣ x_{k} y_{k} - 0 ∣ < 1$ for all large $k$ ; but $∣ x_{k} y_{k} - 0 ∣ = ∣ 1 ∣ = 1$ by [F8] and [F11], and $1 < 1$ is false by [F10]. So no threshold satisfies the test: the product does not converge to $0$ .

3.1step 1.1step 1.2step 2.1F7F13∎

The refutation is complete: $(x_{k})$ is null by step 1.2, $(y_{k})$ diverges to $+ \infty$ by step 1.1, and the product fails to converge to $0$ by step 2.1. Moreover $(y_{k})$ is not bounded in the sense of [F13]: if $∣ y_{k} ∣ \leq M$ held for every $k$ , then for an index beyond the threshold belonging to the real $M$ of step 1.1 one would have both $y_{k} > M$ and $y_{k} \leq ∣ y_{k} ∣ \leq M$ . Since [F7] is true under the additional boundedness hypothesis with the same conclusion, the hypothesis is exactly what the counterexample removes, and it cannot be dropped.

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The quotient rule needs a nonzero limit in the denominator: $1 / (k + 1)^{- 1} = k + 1$

Statement refuted

Refuted claim: if $x_{k} \to x$ and $y_{k} \to y$ and $x_{k} \neq 0$ for every $k$ , then $y_{k} / x_{k}$ converges.

The witnesses are $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = 1$ : both converge — to $0$ and to $1$ respectively — and every $x_{k}$ is nonzero, but $y_{k} / x_{k} = k + 1$ diverges to $+ \infty$ . The quotient clause of Algebra of limits: sums, scalar multiples, products and quotients requires the limit of the denominator to be nonzero, and this is the extremal failure of exactly that hypothesis. It also instantiates the reciprocal bridge of For positive terms, null and divergence to $+ \infty$ are reciprocal.

Facts & Assumptions

Given: The sequences $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = 1$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F1]

$(x_{k})$ converges to $0$ and every $x_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Reciprocal bridge: for a sequence $(z_{k})$ with $z_{k} > 0$ for every $k$ , $(z_{k})$ converges to $0$ if and only if $(1 / z_{k})$ diverges to $+ \infty$ (For positive terms, null and divergence to $+ \infty$ are reciprocal).

[F3]

Divergence to $+ \infty$ : $z_{k} \to + \infty$ means that for every real $M$ there is $K$ with $z_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F4]

Field arithmetic: every nonzero $u$ has a multiplicative inverse with $u \cdot u^{- 1} = 1$ and $(u^{- 1})^{- 1} = u$ , multiplication is commutative, and $u \cdot 1 = u$ (Field).

[F5]

A sequence diverging to $+ \infty$ is unbounded, and an unbounded sequence is not convergent, the latter being the contrapositive of "convergent implies bounded" (Every convergent sequence is bounded, Divergence to $+ \infty$ and to $- \infty$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F6]

Quotient clause of the algebra of limits: if $(p_{k})$ converges to $p$ , $(q_{k})$ converges to $q$ , and $q \neq 0$ with $q_{k} \neq 0$ for every $k$ , then $p_{k} / q_{k}$ converges to $p / q$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F7]

Every constant real sequence converges to its value: $∣ c - c ∣ = 0 < ε$ at every index for every rational $ε > 0$ (Limits and Cauchy sequences of reals).

Counterexample

technique · direct

1.1givenF1F7

The sequence $(x_{k})$ is positive termwise and converges to $0$ by [F1]; the constant sequence $(y_{k}) = (1)$ converges to $1$ by [F7].

2.1step 1.1F2F3

By the forward direction of the reciprocal bridge [F2], applied to the positive null sequence $(x_{k})$ of step 1.1, the sequence $(1 / x_{k})$ diverges to $+ \infty$ in the sense of [F3].

2.2step 1.1F4

At every index the quotient equals the reciprocal: $y_{k} / x_{k} = 1 \cdot x_{k}^{- 1} = x_{k}^{- 1} = ((k + 1)^{- 1})^{- 1} = k + 1 = 1 / x_{k}$ , using $u \cdot 1 = u$ and the involution $(u^{- 1})^{- 1} = u$ of [F4] at the nonzero element $u = k + 1$ .

3.1step 1.1step 2.1step 2.2F5F6∎

Therefore $(y_{k} / x_{k}) = (1 / x_{k})$ diverges to $+ \infty$ by step 2.1, so by [F5] the quotient is unbounded and not convergent: the refuted claim fails. The quotient clause [F6] is not contradicted — its denominator-limit hypothesis $q \neq 0$ fails here, since the limit of the denominator is $x = 0$ by step 1.1, while its other hypotheses ( $x_{k} \neq 0$ termwise) do hold.

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The squeeze theorem needs the two bounds to share their limit

Statement refuted

Refuted claim: if $a_{k} \leq x_{k} \leq b_{k}$ eventually and $(a_{k})$ and $(b_{k})$ both converge, then $(x_{k})$ converges.

The witnesses are the constant sequences $a_{k} : = - 1$ and $b_{k} : = 1$ together with the alternating sequence $x_{k} : = s_{k}$ of FALSE: every bounded sequence converges; the inequalities hold at every index, the two bounds converge to $- 1$ and $1$ respectively, and $(s_{k})$ does not converge. So The squeeze theorem genuinely needs the two limit values to coincide, not merely the two bounds to exist.

Facts & Assumptions

Given: The unique sequences $(s_{k})$ of reals with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ for every $k$ (The recursion theorem), and the constant sequences $a_{k} : = - 1$ , $b_{k} : = 1$ , $x_{k} : = s_{k}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F1]

The sequence $(s_{k})$ is the alternating sequence of FALSE: every bounded sequence converges: it is introduced there by the recursion $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ (FALSE: every bounded sequence converges).

[F2]

$(s_{k})$ does not converge (FALSE: every bounded sequence converges).

[F3]

Absolute value is even: $∣ - u ∣ = ∣ u ∣$ (Basic properties of the absolute value).

[F4]

Induction principle: if $P (0)$ holds and $P (k)$ implies $P (σ (k))$ for every $k$ , then $P (k)$ holds for every natural $k$ (The principle of mathematical induction).

[F5]

Absolute value: $∣ u ∣ = u$ if $u \geq 0$ and $∣ u ∣ = - u$ if $u < 0$ ; so $∣ 0 ∣ = 0$ , $∣ 1 ∣ = 1$ and, for any $u$ , the equality $∣ u ∣ = 1$ forces $u = 1$ when $u \geq 0$ and $u = - 1$ when $u < 0$ (Absolute value in an ordered field).

[F6]

Trichotomy: for all reals $u, v$ exactly one of $u < v$ , $u = v$ , $v < u$ holds, and the order is transitive (Ordered field).

[F7]

$0 < 1$ , so by adding constants $- 1 < 1$ and in particular $1 \neq - 1$ , while $- 1 \leq 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities).

[F8]

Convergence: $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F9]

Squeeze theorem: if $a_{k} \leq x_{k} \leq b_{k}$ eventually and $(a_{k})$ , $(b_{k})$ both converge to the same real $L$ , then $(x_{k})$ converges to $L$ (The squeeze theorem).

Counterexample

technique · direct

1.1givenF1F3F4F5

By induction [F4], $∣ s_{k} ∣ = 1$ for every $k$ : the base is $∣ s_{0} ∣ = ∣ 1 ∣ = 1$ by [F5], and if $∣ s_{k} ∣ = 1$ then $∣ s_{σ (k)} ∣ = ∣ - s_{k} ∣ = ∣ s_{k} ∣ = 1$ by [F3] and the recursion [F1].

1.2F5F8

The constant sequences converge to their values: given a rational $ε > 0$ , the estimates $∣ - 1 - (- 1) ∣ = ∣ 0 ∣ = 0 < ε$ and $∣ 1 - 1 ∣ = ∣ 0 ∣ = 0 < ε$ hold at every index by [F5], so the threshold $K = 0$ witnesses the condition of [F8] in each case.

1.3F2F7

$(s_{k})$ does not converge by [F2], and the two limits $- 1$ and $1$ are distinct by [F7].

2.1step 1.1F5F6F7

Consequently $s_{k} \in {1, - 1}$ for every $k$ : if $s_{k} \geq 0$ then $s_{k} = ∣ s_{k} ∣ = 1$ by [F5], while if $s_{k} < 0$ then $s_{k} = - ∣ s_{k} ∣ = - 1$ by [F5]; trichotomy [F6] makes the two cases exhaustive. In both cases $- 1 \leq s_{k} \leq 1$ by [F7].

3.1step 1.2step 1.3step 2.1F9∎

Steps 2.1 and 1.2 give $a_{k} \leq x_{k} \leq b_{k}$ at every index (hence eventually) with $(a_{k}) \to - 1$ and $(b_{k}) \to 1$ , while by step 1.3 the middle sequence does not converge; since the two bound limits differ, the equal-limit hypothesis of [F9] fails and the squeeze gives no conclusion — and indeed the conclusion is false.

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Geometric sequences: the complete classification for every real ratio

Example

For every real $r$ , let $a_{k} = r^{k}$ for natural $k \geq 0$ , using the integer powers of Integer powers $a^{m}$ , so $a_{0} = 1$ even when $r = 0$ , and $a_{k + 1} = a_{k} r$ . This notation is valid for negative real ratios and is not rational exponentiation of a negative base. The complete classification is: if $∣ r ∣ < 1$ , then $r^{k} \to 0$ ; if $r = 1$ , the sequence is constantly $1$ and converges to $1$ ; if $r = - 1$ , its even and odd subsequences are constantly $1$ and $- 1$ , so it is bounded but divergent; if $r > 1$ , then $r^{k} \to + \infty$ ; if $r < - 1$ , then $∣ r^{k} ∣ \to + \infty$ , while the even subsequence diverges to $+ \infty$ and the odd subsequence to $- \infty$ , and the whole sequence neither converges nor diverges to either signed infinity. Consequently $(r^{k})$ converges exactly when $- 1 < r \leq 1$ . Here signed-infinity assertions mean Divergence to $+ \infty$ and to $- \infty$ , not a finite limit value. Quantitative checks: if $0 < ∣ r ∣ < 1$ and $h = ∣ r ∣^{- 1} - 1 > 0$ , then $∣ r^{k} ∣ \leq (1 + k h)^{- 1}$ for every natural $k$ , and for $k \geq 1$ one has $∣ r^{k} ∣ < (k h)^{- 1}$ ; if $∣ r ∣ > 1$ and $d = ∣ r ∣ - 1 > 0$ , then $∣ r^{k} ∣ \geq 1 + k d$ . The weak Bernoulli estimates have equality for $k = 0, 1$ and are strict for $k \geq 2$ . Thus $(1 / 2)^{k}$ and $(- 1 / 2)^{k}$ are both null, $2^{k}$ diverges positively, and $(- 2)^{k}$ grows in absolute value while alternating sign.

Facts & Assumptions

Given: A real $r$ , the sequence $a_{k} : = r^{k}$ with integer powers (Integer powers $a^{m}$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences), the alternating sequence $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ and its index map $n$ with $n_{0} = 0$ , $n_{σ (j)} = σ (σ (n_{j}))$ (FALSE: every bounded sequence converges), and $m_{j} : = σ (n_{j})$ .

[F1]

Integer powers: $a^{0} = 1$ and $a^{j + 1} = a^{j} a$ for every real $a$ and every natural $j$ (Integer powers $a^{m}$ ).

[F2]

Power laws: $a^{i + j} = a^{i} a^{j}$ , $(a^{i})^{j} = a^{i j}$ and $(a b)^{j} = a^{j} b^{j}$ for all naturals $i, j$ ; and if $a \neq 0$ then $a^{j} \neq 0$ for every $j$ (Laws of integer exponents).

[F3]

Signs of powers: if $a \geq 0$ then $a^{j} \geq 0$ , and if $a > 0$ then $a^{j} > 0$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[F4]

Bernoulli's inequality: for $x \geq - 1$ and every natural $j$ , $(1 + x)^{j} \geq 1 + j x$ , with equality if and only if $j \leq 1$ or $x = 0$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[F5]

Archimedean property: for every real $x$ there is a natural number $j \geq 1$ with $x < j \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F6]

Inverses of positives: $a > 0$ implies $a^{- 1} > 0$ ; and $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order).

[F7]

Canonical naturals: $n \mapsto n \cdot 1_{R}$ is strictly increasing, hence preserves non-strict order, on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[F8]

$0 < 1$ and $1 \neq 0$ (The multiplicative identity is positive).

[F9]

Absolute value: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ - u ∣ = ∣ u ∣$ , $∣ u ∣ \geq 0$ , $∣ u ∣ = 0$ if and only if $u = 0$ , and $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[F10]

A product with a zero factor vanishes: $0 \cdot a = 0$ (Multiplication by zero: $0 \cdot a = 0$ ).

[F11]

The alternating sequence: $(s_{k})$ is the unique sequence with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ (FALSE: every bounded sequence converges).

[F25]

The index map $n$ of the same item satisfies $n_{0} = 0$ , $n_{σ (j)} = σ (σ (n_{j}))$ , is strictly increasing, and has $s_{n_{j}} = 1$ for every $j$ (FALSE: every bounded sequence converges).

[F12]

Growth of index maps: if $p$ is strictly increasing then $p_{j} \geq j$ for every $j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[F13]

Consecutive comparisons suffice: if $p_{i} < p_{σ (i)}$ for every $i$ then $p$ is strictly increasing (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[F14]

Discreteness: for naturals $a, b$ , $a < b ⟺ σ (a) \leq b$ (Discreteness: $σ (n)$ is the immediate successor).

[F15]

Subsequences inherit limits, and the divergence test: if a sequence converges to $x$ and $p$ is strictly increasing, then the subsequence along $p$ converges to $x$ ; consequently, if two subsequences converge to different limits the sequence does not converge (Subsequences inherit the limit).

[F16]

Every convergent sequence is bounded (Every convergent sequence is bounded).

[F17]

Divergence to $+ \infty$ and to $- \infty$ means, respectively, that for every real $M$ one has $x_{k} > M$ eventually and $x_{k} < M$ eventually (Divergence to $+ \infty$ and to $- \infty$ ).

[F18]

Trichotomy, and $u \leq v < w$ implies $u < w$ (Ordered field).

[F19]

Scaling a strict inequality by a positive factor: if $c > 0$ then $a < b ⟺ a c < b c$ (Sign rules for products and monotonicity of multiplication).

[F20]

Convergence: $(z_{k})$ converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F21]

Field arithmetic: multiplication is associative and commutative, it distributes over addition, and every nonzero element has a multiplicative inverse (Field).

[F22]

Adding a constant preserves the strict order: $a < b$ implies $a + c < b + c$ (Order is preserved by adding a constant and by adding inequalities).

[F23]

Induction principle: if a property $P$ satisfies $P (0)$ and $P (n) \Rightarrow P (σ (n))$ for all $n$ , then $P (n)$ holds for all $n \in N$ (The principle of mathematical induction).

[F24]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ ; it is unbounded otherwise (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Verification

technique · cases

1.1baseihdischarge-inductiongivenF1F8F9F23

Absolute values of powers: for every real $r$ and every $k$ one has $∣ r^{k} ∣ = ∣ r ∣^{k}$ , by induction. The base case is $∣ r^{0} ∣ = ∣ 1 ∣ = 1 = ∣ r ∣^{0}$ by [F1], [F8] and [F9]; and assuming $∣ r^{k} ∣ = ∣ r ∣^{k}$ for a fixed $k$ , the recursion [F1] and multiplicativity [F9] give $∣ r^{σ (k)} ∣ = ∣ r^{k} r ∣ = ∣ r^{k} ∣ ∣ r ∣ = ∣ r ∣^{k} ∣ r ∣ = ∣ r ∣^{σ (k)}$ . Hence by the induction principle [F23] the identity holds for every $k$ ; in particular $∣ (- b)^{k} ∣ = ∣ - b ∣^{k} = b^{k}$ for every real $b \geq 0$ and every $k$ .

1.2givenF4F5F6F7F9F17F19algebra

Growth for bases above $1$ : let $b > 1$ be real and put $d : = b - 1 > 0$ . Bernoulli [F4] gives $b^{k} = (1 + d)^{k} \geq 1 + k d$ for every $k$ , with equality exactly when $k \leq 1$ , since $d \neq 0$ . Given any real $M$ , apply [F5] to the real $1 + ∣ M / d ∣$ to get a natural $N \geq 1$ with $N > 1 + ∣ M / d ∣$ ; then $N d > d + ∣ M ∣ \geq ∣ M ∣$ , so for every $k \geq N$ the order preservation [F7] and scaling by $d > 0$ [F19] give $k d \geq N d > ∣ M ∣ \geq M$ , hence $b^{k} \geq 1 + k d > M$ . Therefore $b^{k} \to + \infty$ in the sense of [F17].

1.3assume-case zerogivenF1F9F10F20

The case $r = 0$ : here $r^{0} = 1$ by [F1], and $r^{k} = 0$ for every $k \geq 1$ by induction, since $r^{σ (k)} = r^{k} r$ and a product with a zero factor vanishes [F10]. Given a rational $ε > 0$ , for every $k \geq 1$ we have $∣ r^{k} - 0 ∣ = ∣ 0 ∣ = 0 < ε$ by [F9], so $r^{k} \to 0$ by [F20] with threshold $K = 1$ .

1.4F9F20

Constant sequences converge to their values: for every real $c$ , given a rational $ε > 0$ one has $∣ c - c ∣ = ∣ 0 ∣ = 0 < ε$ at every index by [F9], so the threshold $K = 0$ witnesses [F20].

1.5baseihdischarge-inductiongivenF1F2F9F11F21F23

Alternating powers: for every positive real $b$ and every $k$ , $(- b)^{k} = s_{k} b^{k}$ by induction: the base case is $(- b)^{0} = 1 = s_{0} b^{0}$ , and if $(- b)^{k} = s_{k} b^{k}$ then $(- b)^{σ (k)} = (- b)^{k} (- b) = s_{k} b^{k} (- b) = (- s_{k}) b^{k} b = s_{σ (k)} b^{σ (k)}$ by [F1], [F2], [F11] and field arithmetic [F21]. Alongside, $∣ s_{k} ∣ = 1$ for every $k$ by induction: $∣ s_{0} ∣ = ∣ 1 ∣ = 1$ , and $∣ s_{σ (k)} ∣ = ∣ - s_{k} ∣ = ∣ s_{k} ∣ = 1$ by [F9] and [F11].

1.6givenF11F12F13F14F25

The index maps: $n$ is strictly increasing with $n_{0} = 0$ and $s_{n_{j}} = 1$ [F25], so $n_{j} \geq j$ [F12]; the map $m_{j} = σ (n_{j})$ is strictly increasing, because $n_{j} < n_{σ (j)}$ gives $m_{j} = σ (n_{j}) \leq n_{σ (j)} < σ (n_{σ (j)}) = m_{σ (j)}$ by [F14] and [F13], and then $m_{j} \geq j$ by [F12]; moreover $s_{m_{j}} = s_{σ (n_{j})} = - s_{n_{j}} = - 1$ by [F11].

2.1assume-case smallstep 1.1givenF2F3F4F5F6F7F9F19F20F21algebra

The case $0 < ∣ r ∣ < 1$ : put $b : = ∣ r ∣ \in (0, 1)$ and $h : = b^{- 1} - 1 > 0$ , positive by [F6]. Bernoulli [F4] applied to $h \geq - 1$ gives $(b^{- 1})^{k} = (1 + h)^{k} \geq 1 + k h$ for every $k$ , with equality exactly when $k \leq 1$ ; since $b^{k} (b^{- 1})^{k} = (b b^{- 1})^{k} = 1^{k} = 1$ by [F2], uniqueness of inverses [F21] gives $b^{k} = ((b^{- 1})^{k})^{- 1} \leq (1 + k h)^{- 1}$ by the reciprocal reversal [F6], both sides positive by [F3]; for $k \geq 1$ also $1 + k h > k h > 0$ , so $(1 + k h)^{- 1} < (k h)^{- 1}$ and $∣ r^{k} ∣ = b^{k} < (k h)^{- 1}$ by step 1.1. Now let a rational $ε > 0$ be given; [F5] applied to $1 / (h ε)$ supplies a natural $N \geq 1$ with $N > 1 / (h ε)$ , so $(N h)^{- 1} < ε$ ; for every $k \geq N$ the order preservation [F7] and scaling by $h > 0$ [F19] give $k h \geq N h$ , hence $(k h)^{- 1} \leq (N h)^{- 1} < ε$ by [F6], and therefore $∣ r^{k} - 0 ∣ = ∣ r^{k} ∣ < ε$ by [F9]. So $r^{k} \to 0$ .

2.2assume-case unitstep 1.4givenF1

The case $r = 1$ : then $r^{k} = 1$ for every $k$ , by induction from $1^{0} = 1$ and $1^{σ (k)} = 1^{k} \cdot 1 = 1$ by [F1]; hence $r^{k} \to 1$ by step 1.4.

2.3assume-case negunitstep 1.1step 1.4step 1.5step 1.6givenF8F15F17F18F22F24

The case $r = - 1$ : by step 1.5 with $b = 1$ one has $r^{k} = (- 1)^{k} = s_{k}$ for every $k$ . The even subsequence along $n$ is constantly $1$ and the odd subsequence along $m$ is constantly $- 1$ : $r^{n_{j}} = s_{n_{j}} = 1$ and $r^{m_{j}} = s_{m_{j}} = - 1$ by step 1.6, and both index maps are strictly increasing by step 1.6, so these are subsequences in the sense of [F15] converging to $1$ and to $- 1$ respectively. The limits differ: $- 1 < 1$ since $0 < 1$ by [F8] and adding $- 1$ preserves the order by [F22], so $1 \neq - 1$ by [F18], and the divergence test [F15] shows that $(r^{k})$ does not converge. It is bounded in the sense of [F24], since $∣ r^{k} ∣ = ∣ r ∣^{k} = 1$ for every $k$ by step 1.1, and it diverges to neither signed infinity: at the bound $M = 1$ the condition $x_{k} > 1$ of [F17] fails at every index, and at $M = - 1$ the condition $x_{k} < - 1$ fails at every index.

2.4assume-case bigstep 1.2givenF3F16F17F24

The case $r > 1$ : with $b : = r > 1$ , step 1.2 gives $b^{k} \to + \infty$ , and $r^{k} = b^{k}$ for every $k$ ; hence $r^{k} \to + \infty$ in the sense of [F17]. Since the terms are positive and eventually exceed every real bound, the sequence is unbounded and cannot converge by [F16].

2.5assume-case negbigstep 1.1step 1.2step 1.5step 1.6givenF9F16F17F24

The case $r < - 1$ : put $b : = ∣ r ∣ > 1$ . By step 1.5, $r^{k} = s_{k} b^{k}$ for every $k$ , so $∣ r^{k} ∣ = ∣ s_{k} ∣ b^{k} = b^{k} \to + \infty$ by step 1.1, step 1.2 and [F9]; since $b^{k}$ exceeds every real eventually, no real $M$ satisfies $∣ r^{k} ∣ \leq M$ for all $k$ , so $(r^{k})$ is unbounded in the sense of [F24] and therefore not convergent by [F16]. Along the even index map, $r^{n_{j}} = s_{n_{j}} b^{n_{j}} = b^{n_{j}}$ by step 1.6: given a real $M$ , step 1.2 gives $K$ with $b^{k} > M$ for all $k \geq K$ , while $n_{j} \geq j \geq K$ for $j \geq K$ by step 1.6, so $r^{n_{j}} = b^{n_{j}} > M$ for all $j \geq K$ , which is divergence of that subsequence to $+ \infty$ by [F17]. Along the odd index map, $r^{m_{j}} = s_{m_{j}} b^{m_{j}} = - b^{m_{j}}$ by step 1.6: given a real $M$ , step 1.2 supplies $K$ such that $b^{k} > - M$ for $k \geq K$ , and $m_{j} \geq j \geq K$ for $j \geq K$ by step 1.6; hence $b^{m_{j}} > - M$ , so $- b^{m_{j}} < M$ for every $j \geq K$ , so the odd subsequence diverges to $- \infty$ by [F17]. Finally the whole sequence fails both signed-infinity tests at $M = 0$ : the condition $x_{k} > 0$ eventually fails because $r^{m_{j}} = - b^{m_{j}} < 0$ at the arbitrarily large indices $m_{j} \geq j$ , and the condition $x_{k} < 0$ eventually fails because $r^{n_{j}} = b^{n_{j}} > 0$ at the arbitrarily large indices $n_{j} \geq j$ ; so it neither converges nor diverges to either signed infinity.

3.1cases-exhaustivestep 1.2step 1.3step 2.1step 2.2step 2.3step 2.4step 2.5givenF18∎

The cases are exhaustive: by trichotomy [F18] every real $r$ satisfies exactly one of $r < - 1$ , $r = - 1$ , $- 1 < r < 1$ , $r = 1$ , $r > 1$ , and these have been disposed of in steps 2.5, 2.3, 2.1 and 1.3 (for $- 1 < r < 1$ , splitting at $r = 0$ ), 2.2 and 2.4 respectively. So $(r^{k})$ converges exactly when $- 1 < r \leq 1$ : to $0$ for $- 1 < r < 1$ by steps 1.3 and 2.1, and to $1$ for $r = 1$ by step 2.2, while it diverges for $r = - 1$ , for $r > 1$ and for $r < - 1$ . The quantitative bounds are those derived in steps 1.2 and 2.1 — $∣ r^{k} ∣ \leq (1 + k h)^{- 1}$ and $∣ r^{k} ∣ < (k h)^{- 1}$ for $k \geq 1$ when $0 < ∣ r ∣ < 1$ , and $∣ r^{k} ∣ \geq 1 + k d$ when $∣ r ∣ > 1$ — and the displayed numerical instances $(1 / 2)^{k}$ , $(- 1 / 2)^{k}$ (null), $2^{k}$ (diverging to $+ \infty$ ) and $(- 2)^{k}$ (absolute values diverging, signs alternating) are the substitutions $r = \pm 1 / 2$ and $r = \pm 2$ into the proved clauses.

Sequences and Limits: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The sequence 1/(k+1)1/(k+1) converges to 00

Example

Facts & Assumptions

Verification

A rational expression in kk: (2(k+1)+3)/(5(k+1)+1)→2/5(2(k+1)+3)/(5(k+1)+1)→2/5

Example

Facts & Assumptions

Verification

The alternating sequence (−1)k/(k+1)(−1)k/(k+1) converges to 00

Example

Facts & Assumptions

Verification

The square roots kk​ diverge to +∞+∞

Example

Facts & Assumptions

Verification

The successive differences k+1−kk+1​−k​ converge to 00

Example

Facts & Assumptions

Verification

Changing the first hundred terms leaves convergence to 00 and boundedness intact

Example

Facts & Assumptions

Verification

Two squeeze computations: 1/((k+1)k+1)→01/((k+1)k+1​)→0 and (k+(−1)k)/(k+1)→1(k+(−1)k)/(k+1)→1

Example

Facts & Assumptions

Verification

Even and odd subsequences of 1/(k+1)1/(k+1) converge to 00

Example

Facts & Assumptions

Verification

The absolute value of the limit, and the limit of the absolute value

Example

Facts & Assumptions

Verification

The sum of two divergent sequences can converge: k+(−k)k+(−k)

Statement refuted

Facts & Assumptions

Counterexample

A null sequence times an unbounded sequence need not be null: (k+1)−1⋅(k+1)=1(k+1)−1⋅(k+1)=1

Statement refuted

Facts & Assumptions

Counterexample

The quotient rule needs a nonzero limit in the denominator: 1/(k+1)−1=k+11/(k+1)−1=k+1

Statement refuted

Facts & Assumptions

Counterexample

The squeeze theorem needs the two bounds to share their limit

Statement refuted

Facts & Assumptions

Counterexample

Geometric sequences: the complete classification for every real ratio

Example

Facts & Assumptions

Verification

Sources

The sequence $1 / (k + 1)$ converges to $0$

A rational expression in $k$ : $(2 (k + 1) + 3) / (5 (k + 1) + 1) \to 2 / 5$

The alternating sequence $(- 1)^{k} / (k + 1)$ converges to $0$

The square roots $\sqrt{k}$ diverge to $+ \infty$

The successive differences $\sqrt{k + 1} - \sqrt{k}$ converge to $0$

Changing the first hundred terms leaves convergence to $0$ and boundedness intact

Two squeeze computations: $1 / ((k + 1) \sqrt{k + 1}) \to 0$ and $(k + (- 1)^{k}) / (k + 1) \to 1$

Even and odd subsequences of $1 / (k + 1)$ converge to $0$

The sum of two divergent sequences can converge: $k + (- k)$

A null sequence times an unbounded sequence need not be null: $(k + 1)^{- 1} \cdot (k + 1) = 1$

The quotient rule needs a nonzero limit in the denominator: $1 / (k + 1)^{- 1} = k + 1$