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