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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.

Depends on

Used by

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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)

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