# $\limsup s_n = \infty$, $\liminf t_n >0$, prove that $\limsup s_n t_n = \infty$

I need some help with the last step. Here's what I have already done.

Proof:

$\limsup s_n =0$ implies that $\lim_{N\to\infty}\sup\{s_n:n>N\}=\infty$, by definition. We know that $\sup\{s_n:n>N\}$ itself is a decreasing bounded sequence. So its limit equals to its infimum. So we have $\inf_{N}\{\sup\{s_n:n>N\}\}=\infty$. By definition of the infimum, we see that $\sup\{s_n:n>N\}=\infty \ \ \forall N$.

Similarly, since $\lim_{N \to \infty} \inf\{t_n:n>N\}>0.$ Since $\inf\{t_n:n>N\}$ itself is an increasing bounded sequence, so its supremum must be larger than 0, too. So $\sup\{\inf\{t_n:n>N\}\}>0$. But this implies that $0$ is NOT an upperbound, so there must be some $N_0$ such that $\inf\{t_n:n>N_0\}>0$. So if $n>N_0$, then $t_n>0$.

Any idea on how to continue? I am not sure how should I collect what I have done and try to relate that to $\sup\{s_nt_n:n>N\}$.

I'm not sure if what you've done so far is useful. To prove this statement, you simply have to formalize the idea that $(t_n)$ is always above some $\varepsilon > 0$ for large enough $n$, and so since $(s_n)$ is never bounded above, for any upper bound $B$ you can always find some $N$ such that $s_N > \frac{B}{\varepsilon}$ and so $s_Nt_N > B$.
This means that no $B$ can be an upper bound for $(s_nt_n)$, and your result follows.