Does a sequence $a_n$ converge if $|a_n-\frac{1}{n} \sum_{i=1}^n a_i| \to 0$, as $n \to \infty$? Let $\{a_n\}$ be a real sequence. If $\lim\limits_{n\to \infty} \left|a_n - \frac{1}{n}\sum\limits_{i=1}^n a_i \right|= 0$, do we have $\lim\limits_{n\to \infty} a_n$ convergent?
This is somehow an inverse version of Cesaro mean convergence, which says that if  $\lim\limits_{n\to \infty} a_n = L$, then $\lim\limits_{n\to \infty} \frac{1}{n}\sum\limits_{i=1}^n a_i = L$. See 

On cesaro convergence: If $ x_n \to x $ then $ z_n = \frac{x_1 + \dots +x_n}{n} \to x $

 A: I first tried
$a_n = \ln(n)$.
$\frac{1}{n}\sum_{i=1}^n a_i
=\ln(n!)/n
\approx (n\ln(n)-n+O(\ln(n)))/n
=\ln(n)-1+o(1)
$.
Close.
Looking at something
slower growing,
I tried
$a_n = \ln\ln(n+1)$
and this works.
$\int \ln(\ln(x))dx
=x\ln(\ln(x))+\int \dfrac{dx}{\ln(x)}
=x\ln(\ln(x))+\dfrac{x}{\ln(x)}+O(\dfrac{x}{\ln^2(x)})
$
By Euler-Maclaurin,
$\begin{array}\\
\sum_{k=2}^n \ln\ln(k)
&=\int_2^n \ln\ln(x)dx+\dfrac{\ln\ln(n)+\ln\ln(2)}{2}+O((\ln\ln(n))')\\
&=n\ln\ln(n)+\dfrac{n}{\ln(n)}+O(\dfrac{n}{\ln^2(n)})+\dfrac{\ln\ln(n)+\ln\ln(2)}{2}+O(\dfrac1{n\ln(n)})\\
\text{so}\\
\dfrac{\sum_{k=2}^n \ln\ln(k)}{n}
&=\ln\ln(n)+\dfrac{1}{\ln(n)}+O(\dfrac{1}{\ln^2(n)})+\dfrac{\ln\ln(n)+\ln\ln(2)}{n}+O(\dfrac1{n^2\ln(n)})\\
&=\ln\ln(n)+o(1)\\
\end{array}
$
Therefore
the limit is zero and
$a_n$ diverges.
A: No: consider for example $a_n=\log\log n$.
Edited to add: go upvote marty cohen's answer for working out the details!
A: This is a second answer
because I am
turning the problem completely around.
I will show that
there is a sequence
$a_n$
such that
$\lim_{n\to \infty} \big|a_n - \frac{1}{n}\sum_{i=1}^n a_i \big|= 0
$,
$a_n$ is bounded,
and $a_n$ does not converge.
$\lim_{n\to \infty} \big|a_n - \frac{1}{n}\sum_{i=1}^n a_i \big|= 0
$
means that
$a_n
=\frac{1}{n}\sum_{i=1}^n a_i +f(n)
$
where
$f(n) \to 0$
as
$n \to \infty$.
Then
$na_n
=\sum_{i=1}^n a_i +nf(n)
$
or
$(n-1)a_n
=\sum_{i=1}^{n-1} a_i +nf(n)
$.
Applying the usual trick
when $\sum_{i=1}^n a_i $
appears in a recurrence,
$na_{n+1}
=\sum_{i=1}^{n} a_i +(n+1)f(n+1)
$.
Subtracting,
$\begin{array}\\
na_{n+1}-(n-1)a_n
&=\sum_{i=1}^{n} a_i +(n+1)f(n+1)
-(\sum_{i=1}^{n-1} a_i +nf(n))\\
&=a_n +(n+1)f(n+1) -nf(n)\\
\end{array}
$
so
$\begin{array}\\
na_{n+1}
&=na_n +(n+1)f(n+1) -nf(n)\\
\text{or}\\
a_{n+1}
&=a_n +(1+1/n)f(n+1) -f(n)\\
\end{array}
$
Therefore
$a_{n+1}-a_n
=(1+1/n)f(n+1) -f(n)
=f(n+1) -f(n)+\frac{f(n+1)}{n}
$.
Summing,
$\begin{array}\\
a_n-a_1
&=\sum_{k=1}^{n-1} (a_{k+1}-a_k)\\
&=\sum_{k=1}^{n-1} (f(k+1) -f(k)+\dfrac{f(k+1)}{k})\\
&=f(n)-f(1)+\sum_{k=2}^{n} \dfrac{f(k)}{k-1}\\
\end{array}
$
Now,
by choosing a
$f(k) \to 0$,
we can get a desired $a_n$.
However,
if we want
$a_n \to \infty$,
we must also have
$\sum_{k=2}^{n} \dfrac{f(k)}{k-1}
\to \infty$.
$f(k) = \frac1{k}$
will not work
because the sum converges.
$f(k) = \frac1{\ln(k)}$
works
because 
$\sum_{k=2}^n \dfrac{1}{\ln(k)(k-1)}
\approx \sum_{k=2}^n \dfrac{1}{k\ln(k)}
\approx \int_{k=2}^n \dfrac{dx}{x\ln(x)}
= \ln(\ln(n))
$.
If we want an $a_n$
that doesn't converge
and stays bounded,
we must find a $f(k)$
such that
$\sum_{k=2}^{n} \dfrac{f(k)}{k-1}
$
behaves like that.
Since
$\sum_{k=2}^n \dfrac{1}{\ln(k)(k-1)}
\approx \ln(\ln(n))
$,
for any $n_0$
there is a $n(n_0)$
such that
$\sum_{k=n_0}^{n(n_0)} \dfrac{1}{\ln(k)(k-1)}
\gt 1
$.
An approximate value is
$n(n_0) \approx n_0^e$.
Therefore,
if we choose $f(k)$
to have constant sign
in each interval
$n_0 \le k \lt n(n_0)$,
with the signs alternating
in consecutive intervals,
as $n$ increases
$\sum_{k=2}^{n} \dfrac{(-1)^{g(k)}}{\ln(k)(k-1)}$
for the appropriate $g(k)$
will increase by 1,
then decrease by 1,
and so on,
and will thus not converge
and be bounded.
