Can an unbounded sequence have a convergent cesaro mean? I was wondering if an unbounded sequence may have a convergent cesaro mean ($\frac{1}{n}\sum_{k=1}^n a_n$). I was maybe thinking of $$a_n = (-n)^n$$
as a sequence having a convergent mean, but I might be wrong. Anyways, how would you proceed to prove such an intuition?

Take the following example
$u_{2n}=-\sqrt{n}$
and
$u_{2n+1}=\sqrt{n}$
then
$v_{2n}=0$
and
$v_{2n+1}=\frac{\sqrt{n}}{2n+1}$.
$(u_n)$ is unbounded.
$(v_n)$ goes to $0$.
