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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 $x_{k} \to 0$ .

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)

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