# Prove that $\lim (s_n t_n) =0$ given $\vert t_n \vert \leq M$ and $\lim (s_n) = 0$

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!

• 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. – DonAntonio Jan 10 '14 at 15:18
• 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. – John Hughes Jan 10 '14 at 15:19
• $\epsilon>0$, 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 $\lim s_n = 0$. – user119228 Jan 10 '14 at 15:19
• @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? – William Muenzinger Jan 10 '14 at 17:25
• 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... – DonAntonio Jan 10 '14 at 18:03

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