Simple sequence convergence question... I know the standard solution to the problem that if a sequence $\{s_n\}$ converges, then it's arithmetic mean converges, and to the same limit. But how about this question?:
Assertion: Let $\{\sigma_n\}$ be the arithmetic mean of real-valued sequence $\{s_n\}$. Show that if $ns_n \to 0$, and $\{\sigma_n\}$ converges, then $\{s_n\}$ converges (not necessarily to the same place).
I have three main questions:
1) Is there a real-valued sequence $s_n$ such that $ns_n \to 0$ but $s_n$ does not converge?
2) What is wrong with the following proof of my assertion, other than the fact that it NOWHERE uses the fact that $\sigma_n$ converges:
For all $\epsilon >0$, there exists an $N$ such that $n>N$ implies $|ns_n| < \epsilon$. Thus $|s_n| < \epsilon/n < \epsilon$. Thus $s_n$ converges to $0$. 
Something must be incorrect here, as I have not used an assumption I presume is necessary.
3) What is the correct proof of this assertion?

Your argument is fine. Putting that in another way, we have
$$0\ \le\ |s_n|\ \le \ n\,|s_n|\ =\ |\,n\,s_n\,|\ \to 0$$
so $s_n$ must also converge to $0$.

Anyway, we can get back $s_n$ from $\sigma_n=\displaystyle\frac{s_1+\dots+s_n}n$ as follows: 
$s_1=\sigma_1,\quad s_2=2\sigma_2-\sigma_1,\quad s_3=3\sigma_3-2\sigma_2,\quad\ \  s_n=n\,\sigma_n\ -\ (n-1)\,\sigma_{n-1} $, 
also because $n\,\sigma_n\ =\ s_1+s_2+\dots+s_n$.
So, a correct statement of similar flavor would be

If $(n\,\sigma_n)$ converges, then $s_n\to 0$.

(And also $\sigma_n\to 0$ by your argument.)
