1
$\begingroup$

Prove: If $s_n$ converges to $0$ and $x_n$ is a bounded sequence, then $\underset{n \to \infty}{\lim}(x_ns_n)=0$

I'm have trouble getting started on this proof.

Since I know $s_n$ converges to $0$ I feel like I should start using the epsilon definition with $x_n$

Since $\underset{n \to \infty}{\lim} s_n=0$

So $\left|x_n-s_n\right| \lt \epsilon \implies (=) \quad \left|x_n\right| \lt \epsilon$

But I'm not sure if the aforementioned steps are correct and where to go from here.

$\endgroup$

marked as duplicate by Paramanand Singh, user296602, choco_addicted, R_D, user228113 Jun 18 '16 at 7:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Use the fact there is a $B\gt 0$ such that $|x_n|\lt B$ for all $n$. Then $|x_ns_n-0|\le B|s_n|$. Continue. $\endgroup$ – André Nicolas Jun 18 '16 at 4:15
  • $\begingroup$ @AndréNicolas That looks like an answer to me ... $\endgroup$ – Neal Jun 18 '16 at 4:18
  • $\begingroup$ To the question: You should be looking at the sequence $x_ns_n$< so your inequalities should look like $|x_ns_n - 0|<\epsilon$. $\endgroup$ – Neal Jun 18 '16 at 4:19
  • $\begingroup$ @Neal: For a student new to $\epsilon$-$N$, finishing from there is not necessarily immediate. $\endgroup$ – André Nicolas Jun 18 '16 at 4:21
  • 1
    $\begingroup$ @AndréNicolas this may be a really silly question, but I'm going to go for it anyway. Before I asked this question I studied the theorem and was confused by it. I suppose $B$ is a variable for the boundary, but I don't under where $B|s_n|$ comes from? I recognize $|x_n| \lt B$ because a boundary exist but I don't understand the next step. $\endgroup$ – hax0r_n_code Jun 18 '16 at 4:24
1
$\begingroup$

We write a standard $\epsilon$-$N$ proof. One could alternately use a squeezing argument.

Since the sequence $(x_n)$ is bounded, there is a $B\gt 0$ such that $|x_n|\le B$ for all $n$. It follows that $$|s_nx_n-0|\le B|s_n|\tag{1}$$ for all $n$.

We want to show that given $\epsilon\gt 0$, there is an $N$ such that $|s_nx_n-0|\lt \epsilon$ for all $n\gt N$. By Inequality (1), it is enough to show that there is an $N$ such that if $n\gt N$ then $B|s_n|\lt \epsilon$.

Since the sequence $(s_n)$ has limit $0$, there is an $N$ such that if $n\gt N$ then $|s_n|\lt \frac{\epsilon}{B}$. It follows that if $n\gt N$ then $B|s_n|\lt B\cdot\frac{\epsilon}{B}=\epsilon$, and therefore $|s_nx_n-0|\lt \epsilon$.

$\endgroup$
  • $\begingroup$ I think that's basically equivalent to my proof-but I agree, Andre's proof is a bit more detailed then mine and a better answer. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 5:24
  • $\begingroup$ @Mathemagician1234: There is no need (or use) of Cauchy-Schwarz in your argument, and no need or use of limit of products is product of limits. The sequence $(x_n)$ need not have a limit. $\endgroup$ – André Nicolas Jun 18 '16 at 5:31
  • $\begingroup$ Nicolas It need not have a limit, but the absolute value of the product of the sequences is determined by a factor of M. So yes,the CS inequality can indeed be used in the product limit stage even though ($x_n$) may not itself have a limit. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 17:08
1
$\begingroup$

HINT: If $x_n$ is bounded, then $\exists M>0: |x_n|<M$ for all $n$. Hence $|x_ns_n|<M\varepsilon$ for large $n$. It should be clear that $x_ns_n$ also tends to 0.

$\endgroup$
1
$\begingroup$

There is another simple way if you wish to avoid $\epsilon, \delta$ stuff. You need to use the Squeeze Theorem then.

If $x_{n}$ is bounded then there is a $K > 0$ such that $|x_{n}| < K$ for all $n$. Now we can see that $$0 \leq |x_{n}s_{n}| \leq K|s_{n}|$$ and letting $n \to \infty$ and using Squeeze Theorem we get $$\lim_{n \to \infty}|x_{n}s_{n}| = 0$$ Next we have $$-|x_{n}s_{n}| \leq x_{n}s_{n} \leq |x_{n}s_{n}|$$ and again letting $n \to \infty$ and using Squeeze Theorem we get $$\lim_{n \to \infty}x_{n}s_{n} = 0$$

BTW the proof via $\epsilon,\delta$ is much easier than the above if you understand it and is the preferred approach.

$\endgroup$
  • $\begingroup$ Shoot,I completely forgot about that way to do it. I agree it's a little arduous for the beginner,but the Squeeze Theorem's a good tool to have in your pocket when doing limits. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 5:22
  • $\begingroup$ @Mathemagician1234: I have seen many beginners struggle with $\epsilon$'s and hence try to look for approaches which avoid this. $\endgroup$ – Paramanand Singh Jun 18 '16 at 5:25
  • $\begingroup$ @ParamanandSingh: Squeezing argument is nice here, it has a clearer connection to the intuition. The only advantage of $\epsilon$-$N$ is that it uses no machinery other than the definition of limit. $\endgroup$ – André Nicolas Jun 18 '16 at 5:27
1
$\begingroup$

Since ${x_n}$ is a bounded sequence, then for every $N_1\in \mathbb N$ there is M >0 $\in \mathbb R$ such that when n> $N_1$,$|x_n|$< M.We also know that since $lim_{n\rightarrow\infty} s_n$ =$0$, for every $\epsilon$ >0 $\in \mathbb R$,there is an $N_2\in \mathbb N$ such that when n> $N_2$, $|s_n|$< $\epsilon$. Let $\epsilon$'= M$\epsilon$ and choose n > max {$N_1$,$N_2$}. By the Cauchy Schwarz inequality:

$|x_n s_n|\leq$ $|x_n||s_n|$ < M$\epsilon$ = $\epsilon$'.

So $lim_{n\rightarrow\infty} (x_n s_n)$ = 0.

That does it.

Addendum: Andre Nicolas correctly pointed out that ($x_n$) need not itself have a limit even though it's bounded. While this is very true, the argument above is completely general and refers to the values of the product of the sequence ranges. I've clarified some steps above to show this.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.