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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 $∣ y_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

$(x_{k})$ converges to $x$ when for every rational $ε > 0$ there is $K \in N$ with $∣ x_{k} - x ∣ < \hat{ε}$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[L2]

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

[L3]

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

[L4]

Order arithmetic in $R$ : $0 < 1$ , adding a constant preserves the strict order, and $\leq$ 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 $ε / c > 0$ whenever $ε > 0$ and $c > 0$ (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

[L6]

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

Proof

technique · direct

1.1L2L4

Since $∣ y_{0} ∣ \leq M$ and $∣ y_{0} ∣ \geq 0$ , transitivity gives $M \geq 0$ ; put $C : = M + 1$ , so that $C > 0$ and $∣ y_{k} ∣ \leq M \leq C$ for every $k$ .

2.1step 1.1L5

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

3.1step 2.1L1L2L6

Since $(x_{k})$ converges to $0$ , there is $K \in N$ with $∣ x_{k} ∣ = ∣ x_{k} - 0 ∣ < η$ for every $k \geq K$ .

4.1step 1.1step 3.1L2L3L4

For every $k \geq K$ we get $∣ x_{k} y_{k} - 0 ∣ = ∣ x_{k} ∣ ∣ y_{k} ∣ \leq ∣ x_{k} ∣ C < η C = ε$ , the first inequality from $0 \leq ∣ x_{k} ∣ \leq ∣ x_{k} ∣$ and $0 \leq ∣ y_{k} ∣ \leq C$ , the second from $∣ x_{k} ∣ < η$ and $C > 0$ .

5.1step 4.1L1L6∎

Since the real $ε > 0$ was arbitrary, $(x_{k} y_{k})$ converges to $0$ .

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 $∣ y_{k} ∣ \leq M$ for every $k \geq K$ , 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 $∣ y_{k} ∣ \leq M^{'}$ 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 $x_{k} y_{k} - x y = x_{k} (y_{k} - y) + y (x_{k} - x)$ 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.

A null sequence times a bounded sequence is null

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources