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Let $ (t_n) $ be a bounded sequence, i.e., there exists $ M $ such that $ \vert t_n \vert \leq M $ for all $ n $, and let $ (s_n) $ be a sequence such that $ \lim s_n = 0 $. Prove $ \lim (s_n t_n) =0$.

My attempt:

The given limit, $ \lim (s_n) = 0 $, implies that for any $ \epsilon \in \textbf{R}^+ $ there exists a $ K\in\textbf{R} $ such that for any $ n > K $,

$\vert s_n - 0 \vert < \epsilon $.

Thus if $ \vert s_n \vert = 0 $, then

$ \vert s_n t_n - 0 \vert = 0 < \epsilon $.

Instead suppose $ \vert s_n \vert \neq 0 $. Then since $ 0 \leq \vert t_n \vert \leq M < M + 1 $,

$ \vert s_n t_n - 0\vert \leq \vert s_nM - 0 \vert \leq \vert s_n - 0 \vert M < \vert s_n - 0 \vert (M+1) < \epsilon (M+1) $.

Then since $ \epsilon(M+1) \in \textbf{R}^+ $, it follows that there is a satisfying $ K $.

As this exhausts the cases, $ \lim (s_n t_n) =0$.

Is this proof correct? What are some other ways of proving this? Thanks!

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    $\begingroup$ I can't see why and what for you need the "if $\;|s_n|=0\;$" assumption in the first part. It is completely uneccessary and only makes things more confusing. $\endgroup$ – DonAntonio Jan 10 '14 at 15:18
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    $\begingroup$ You're proof's basically OK (except that "if $|S_n| = 0$ should be "because $lim s_n = 0$" in the 3rd line, as DonAntonio points out). I'd recommend a more positive proof. So say: "Given $\epsilon > 0$, we want to find $L$ with $n > L \implies $|s_n t_n | < \epsilon". Let $\epsilon_1 = \epsilon / M$. Then because $s$ has limit zero, there's a $K$ such that for $n > K$, $|s_n| < \epsilon_1$." And then follow your nose from there. $\endgroup$ – John Hughes Jan 10 '14 at 15:19
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    $\begingroup$ $\epsilon>0$, and $\vert s_n t_n - 0\vert \leq M \vert s_n\vert$ you can not say $\vert s_n t_n - 0\vert \leq \vert s_n M - 0 \vert$. And you can conclude like that : $0 \leq \vert s_n t_n - 0\vert \leq M \vert s_n\vert$ and $ \lim s_n = 0 $. $\endgroup$ – user119228 Jan 10 '14 at 15:19
  • $\begingroup$ @DonAntonio This portion: $ \vert s_n - 0 \vert M < \vert s_n - 0 \vert (M+1) $ required $ \vert s_n \vert = 0 $. I'm not yet using the theorem: $ \lim xy = \lim (x) * \lim(y)$. Is that the source of the problem? $\endgroup$ – William Muenzinger Jan 10 '14 at 17:25
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    $\begingroup$ No @Dargatz, the problem I see is the lack of necessity to check that case. As you can see in my answer, I also didn't use the theorem you mention... $\endgroup$ – DonAntonio Jan 10 '14 at 18:03
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Try the following, perhaps slightly clearer and more direct approach:

We're given $\;|t_n|\le M\;\;\forall\,n\in\Bbb N\;$, and let $\;\epsilon>0\;$ be arbitrary. Since $\;s_n\xrightarrow[n\to\infty]{}0\;$ there exists $\;N\in\Bbb N\;$ s.t. $\;n>N\implies |s_n|<\frac{\epsilon}M\;$ , but then

$$\forall n>N\;:\;\;|s_nt_n|\le|s_n|M<\frac{\epsilon}MM=\epsilon\implies s_nt_n\xrightarrow [n\to\infty]{}0$$

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