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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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Sources

John K. Hunter, An Introduction to Real Analysis (complete author-hosted lecture notes, 307 pp.) (standard reference, not scraped)
Jiri Lebl, Basic Analysis I: Introduction to Real Analysis, Volume I (author-hosted complete book PDF) (standard reference, not scraped)

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