# Proving Convergence

So we're given a sequence $x_n$ and a sequence $y_n$, both of them being of real numbers. We know that $x_n \rightarrow 0$ and that $y_n$ is bounded. We need to prove that $x_n y_n \rightarrow 0$.

My idea was that, since $x_n \rightarrow 0$, multiplying some number of $y_n$ by $0$ would always just be $0$, and we know that we can do this since $y_n$ is bounded (although I don't think it would matter if it were unbounded).

Am I right in thinking this, or is the answer in a completely different direction?

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To understand the importance of having $y_n$ bounded you might try your method with the sequences $x_n = \frac1n$ and $y_n=n$. – Morgan Sherman Oct 19 '12 at 16:01

## 3 Answers

$$|x_ny_n| \leq |x_n| \sup_{n \in \mathbb{N}}|y_n|.$$ If $\varepsilon>0$ is fixed and $N \in \mathbb{N}$ is chosen so that $|x_n|<\varepsilon$ whenever $n>N$, then $|x_n y_n| < \varepsilon \sup_{n \in \mathbb{N}}|y_n|$.

If $\{y_n\}_n$ is unbounded, then the conclusion may be false: $x_n=1/n$ and $y_n=n$ is a counterexample. The idea is that a small number remains small as log as it is multiplied by a number that cannot be arbitrarily large.

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What you’ve suggested doesn’t really make much sense, I’m afraid. You’ve no reason to think that any of the terms $x_n$ is $0$, so in forming the product $x_ny_n$ you’re not in general multiplying $y_n$ by $0$. Go back to the definitions.

1. What does it mean to say that $\langle x_n:n\in\Bbb N\rangle\to 0$?
2. What does it mean to say that $\langle y_n:n\in\Bbb N\rangle$ is bounded?
3. And what does it mean to say that $\langle x_ny_n:n\in\Bbb N\rangle\to 0$?

The answers:

1. For each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|x_n|<\epsilon$ whenever $n\ge n_\epsilon$.
2. There is a real number $M>0$ such that $|y_n|\le M$ for all $n\in\Bbb N$.
3. For each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x_ny_n|<\epsilon$ whenever $n\ge m_\epsilon$.

So now we know what we need to do: we need to show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x_ny_n|<\epsilon$ whenever $n\ge m_\epsilon$. Thus, we need to begin by letting $\epsilon>0$; then we’ll try to use what we’ve been given to find the desired $m_\epsilon$.

We do know that there is some $n_\epsilon$ such that $|x_n|<\epsilon$ whenever $n\ge n_0$; that’s a start on making $|x_ny_n|<\epsilon$. Of course it only works if $|y_n|\le 1$, and we don’t know that that’s true. (It may not be.) We do know, however, that $|y_n|\le M$, so we can at least conclude that $|x_ny_n|\le|x_n|M<\epsilon M$ whenever $n\ge n_0$.

Aha! If we had originally arranged to make $|x_n|<\dfrac{\epsilon}M$, we could now conclude that $$|x_ny_n|\le|x_n|M<\frac{\epsilon}M\cdot M=\epsilon\;.$$ Can we arrange that? Sure: $\dfrac{\epsilon}M>0$, so by hypothesis there is an $n_{\epsilon/M}\in\Bbb N$ such that $|x_n|<\dfrac{\epsilon}M$ whenever $n\ge n_{\epsilon/M}$. Let $m_\epsilon=n_{\epsilon/M}$. Then

$$|x_ny_n|\le|x_n|M<\frac{\epsilon}M\cdot M=\epsilon$$

whenever $n\ge m_\epsilon$. We can do this for any $\epsilon>0$, so by definition $\langle x_ny_n:n\in\Bbb N\rangle\to 0$.

It all starts with the definitions: they tell you what you know and what you need to prove.

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Thank you very much, @Brian. My professor provided scanty details as to even the definitions so that I was mostly working with intuition (which is probably a bad idea). This cleared a lot of things up for me. – Casquibaldo Oct 19 '12 at 16:14
@Casquibaldo: I had a feeling that you might not have seen a detailed argument of this kind; I’m glad that it helped. – Brian M. Scott Oct 19 '12 at 16:17

$\forall \epsilon$, $\exists N$, s.t. $\forall n>N$, $|x_n|<\epsilon$

Also $\exists M,s.t.y_{n}\leq M$ holds forall $n$

So $\forall \epsilon$ $\exists N$ s.t. $\forall n>N$, $|x_n|<\epsilon/M$

thus $\forall n>N$, $|x_n y_n|<\epsilon$ and done

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