TITLE: Does $\lim_{n\to \infty}\sigma_n=0\implies \lim_{n\to \infty} a_n=0$?
QUESTION [6 upvotes]: Let $\{a_n\}$ be a sequence of positive real numbers such that $\lim_{n\to \infty} \sigma_n=0$ where $\sigma_n=\left(\sum_{k=1}^n a_k\right)/n$. 

Does $\lim_{n} \sigma_n=0\implies \lim_n a_n=0$?

I feel that this should be true but I cannot find a proof for this nor can I find a counterexample. I tried my hand at the proof in the following way:
Since $a_n=n\sigma_n-(n-1)\sigma_{n-1},\ n\ge 2$, we have $\lim_n a_n/n=0$. Also, it is not difficult to prove that $\lim\inf_n a_n=0,\ L:=\lim\sup_n\ge 0$. But then I am stuck. 
I will really appreciate short hints, rather than a full answer. Thanks in advance.

REPLY [2 votes]: This is called Cesàro convergence.  For a sequence $a_n$, define
$\sigma_n = \frac{1}{n}\sum_{k=1}^n a_k$.  Then (I) implies (II):
$$
{\rm(I)}\qquad a_n \to L
\\
{\rm(II)}\qquad \sigma_n \to L
$$
But "in general" (II) does not imply (I).  [See the other answers.]  But under "some conditions", (II) does imply (I).  Such a condition may be known as a Tauberian condition.
Ref: https://en.wikipedia.org/wiki/Abelian_and_tauberian_theorems