Let lim $a_n=0$ and $s_N=\sum_{n=1}^{N}a_n$. Show that $\sum a_n$ converges when $\lim_{N\to\infty}s_Ns_{N+1}=p$ for a given $p>0$. Let lim $a_n=0$ and $s_N=\sum_{n=1}^{N}a_n$.
Show that $\sum a_n$ converges when $\lim_{N\to\infty}s_Ns_{N+1}=p$ for a given $p>0$.
I've no idea how to even start. Should I try to prove that $s_N$ is bounded ? 

Put $s_n:=\epsilon_n|s_n|$ with $\epsilon_n\in\{-1,1\}$. Then from
$$\epsilon_n\epsilon_{n+1}|s_n|\>|s_{n+1}|=s_n\>s_{n+1}=:p_n\to p>0\qquad(n\to\infty)$$
it follows that $\epsilon_n=\epsilon_{n+1}$ for $n>n_0$. Assume $\epsilon_n=1$ for all $n> n_0$, the case $\epsilon_n=-1$ being similar.
The equation
$$s_n(s_n+a_{n+1})=s_ns_{n+1}=p_n$$
implies that for all $n$ the quantities $s_n$, $a_{n+1}$, and $p_n$ are related by
$$s_n={1\over2}\left(-a_{n+1}\pm\sqrt{a_{n+1}^2 +4p_n}\right)\ .$$
Since $s_n\geq0$ $\ (n>n_0)$, $\ a_{n+1}\to 0$, $\ p_n\to p>0$ it follows that necessarily
$$s_n={1\over2}\left(-a_{n+1}+\sqrt{a_{n+1}^2 +4p_n}\right)\qquad(n>n_1)\ ,$$
and this implies $\lim_{n\to\infty} s_n=\sqrt{p}$.
