I proved that if a sequence $(s_n)$ converges to a limit $s$ then so does its "average sequence," $(\sigma_n)$ with $\sigma_n=\frac{1}{n}\sum_{k=1}^n s_k$. I found a counterexample for the converse, where $\sigma_n$ converges to a limit but $s_n$ doesn't (a stupid example is $(-1)^n$). Now I am stuck trying to find an example where $\sigma_n$ converges to a (finite) limit, $s_n$ does not, and $s_n$ is unbounded. (Note that we can't have $s_n\rightarrow \infty$ or $s_n\rightarrow -\infty$.)
I've tried $(-1)^nn$, $(-1)^n\lfloor n/2\rfloor$, $\left\lfloor \frac{-1 + \sqrt{8n+1}}{2}\right\rfloor$... but they don't work. I'm stuck. I just can't think of an example. I feel like this is going to smack me in the face.
Can anybody help?
Hint. Work out what $s_n$ needs to be so that $\sigma_n = (-1)^n/\sqrt{n}$.
The reason this seemed reasonable is that for any sudden swing in $\sigma_n$, there needs to be a corresponding value of $s_n$ about $n$ times larger than the swing.