LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

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A null sequence times a bounded sequence is null

Statement

Let $(x_k)$ and $(y_k)$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). If $(x_k)$ converges to $0$ (Limits and Cauchy sequences of reals) and $(y_k)$ is bounded, then the product sequence $(x_k y_k)$ converges to $0$ .

No assumption is made that $(y_k)$ converges. Boundedness alone suffices, and that is why this lemma is stated on its own rather than folded into the product rule for limits.

Facts & Assumptions

Given: Sequences $(x_k)$ and $(y_k)$ of reals with $(x_k)$ converging to $0$ , and a real $M$ with $|y_k| \le M$ for every $k \in \mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

$(x_k)$ converges to $x$ when for every rational $\varepsilon > 0$ there is $K \in \mathbb{N}$ with $|x_k - x| < \hat\varepsilon$ for all $k \ge K$ (Limits and Cauchy sequences of reals).

[L2]

Absolute value: $|uv| = |u|\,|v|$ , $|u| \ge 0$ , and $|u - 0| = |u|$ (Basic properties of the absolute value).

[L3]

Products of inequalities: $0 \le a \le b$ and $0 \le c \le d$ give $ac \le bd$ ; and for $c > 0$ , $a < b$ gives $ac < bc$ (Multiplying inequalities of positives, Sign rules for products and monotonicity of multiplication).

[L4]

Order arithmetic in $\mathbb{R}$ : $0 < 1$ , adding a constant preserves the strict order, and $\le$ and $<$ compose transitively (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field).

[L5]

Inverses: if $c > 0$ then $c^{-1} > 0$ , so $\varepsilon/c > 0$ whenever $\varepsilon > 0$ and $c > 0$ (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

[L6]

Real versus rational $\varepsilon$ : for every real $\eta > 0$ there is a rational $\varepsilon$ with $0 < \varepsilon < \eta$ , by density (The rationals embed densely in the reals) or by the Archimedean property (Every complete ordered field is Archimedean) applied to $1/\eta$ ; consequently the convergence test of Limits and Cauchy sequences of reals may equivalently be run with real $\varepsilon > 0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · direct

1.1

Since $|y_0| \le M$ and $|y_0| \ge 0$ , transitivity gives $M \ge 0$ ; put $C := M + 1$ , so that $C > 0$ and $|y_k| \le M \le C$ for every $k$ .

L2L4

2.1

Let $\varepsilon > 0$ be an arbitrary real and put $\eta := \varepsilon/C$ , which is a real $> 0$ and satisfies $\eta C = \varepsilon$ .

step 1.1L5

3.1

Since $(x_k)$ converges to $0$ , there is $K \in \mathbb{N}$ with $|x_k| = |x_k - 0| < \eta$ for every $k \ge K$ .

step 2.1L1L2L6

4.1

For every $k \ge K$ we get $|x_k y_k - 0| = |x_k|\,|y_k| \le |x_k| \, C < \eta \, C = \varepsilon$ , the first inequality from $0 \le |x_k| \le |x_k|$ and $0 \le |y_k| \le C$ , the second from $|x_k| < \eta$ and $C > 0$ .

step 1.1step 3.1L2L3L4

5.1

Since the real $\varepsilon > 0$ was arbitrary, $(x_k y_k)$ converges to $0$ .

step 4.1L1L6∎

Remarks

The hypothesis on $(y_k)$ looks weaker if it is stated as "eventually bounded", but it is not: a sequence bounded from some index $K$ on is bounded outright. If $|y_k| \le M$ for every $k \ge K$ , then $M' := \max\{\, |y_0|, \dots, |y_{K-1}|, M \,\}$ exists, because a nonempty finite list of reals has a maximum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set), and $|y_k| \le M'$ for every $k$ . That is the same absorption of finitely many initial terms used in Every convergent sequence is bounded.
Boundedness of $(y_k)$ is essential. Take $x_k := 1/(k+1)$ , which is null (that sequence is shown to converge to $0$ in FALSE: limits preserve strict inequalities), and $y_k := (k+1)^2$ . Their product is $x_k y_k = k + 1$ , which is unbounded by the Archimedean property (Every complete ordered field is Archimedean) and hence does not converge at all, since a convergent sequence is bounded (Every convergent sequence is bounded); in particular it is not null. The lemma is therefore sharp in the sense that the bounded factor may not be replaced by an arbitrary one.
The lemma is the workhorse of the product rule: the identity $x_k y_k - xy = x_k(y_k - y) + y(x_k - x)$ exhibits the error as a sum of two products of a null sequence with a bounded one, and boundedness of $(x_k)$ comes from Every convergent sequence is bounded.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 51 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Reed College Math 112, §7.5 Theorems About Convergent Sequences (standard reference, not scraped)
Limit of a sequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.1 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)