$\sum _{1}^{\infty}a_n^2/n$ converges then $(1/N)\sum_{1}^{N}a_n$ converges to zero. Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that 

If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero. 

A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.
Could anyone please help me to prove it?
 A: Fix any $N_0$ and write:
\begin{align*}
\frac{1}{N} \sum\limits_{n = 1}^N |a_n| &= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N |a_n| }{N} \\
&= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N \frac{|a_n|}{\sqrt{n}} \cdot \sqrt{n} }{N} \\
&\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n}  \right) \left(\sum\limits_{n = N_0 + 1}^N n \right) }}{N} \\
&\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} +  \sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n}  \right)}.
\end{align*}
Here on the third line I use CS inequality. To obtain the forth line I simply bound $\sum\limits_{n = N_0 + 1}^N n$ from above by $N^2$. 
Now by taking $N\to\infty$ we get
$$\limsup\limits_{N\to\infty} \frac{1}{N} \sum\limits_{n = 1}^N |a_n| \le  \sqrt{\left(\sum\limits_{n = N_0 + 1}^\infty \frac{a_n^2}{n}  \right)}. $$
As $N_0$ can chosen arbitrarily, the right hand side can be made arbitrarily small --- here we use the fact that $\sum\limits_{n =  1}^\infty \frac{a_n^2}{n} $ is convergent.
A: We use the Cauchy-Schwartz inequality to write
$$ \frac{1}{N}\sum_{n=1}^{N}a_n = 
\sum_{n=1}^{N} \frac{\sqrt{n}}{N}\frac{a_n}{\sqrt{n}} 
\leq \sqrt{\sum_{n=1}^N \frac{n}{N^2} \sum_{n=1}^{N} \frac{a^2_n}{n} }
= \sqrt{\frac{1}{N^2}\frac{N(N+1)}{2}\sum_{n=1}^{N}\frac{a_n^2}{n}}$$
Which converges as $N \to \infty$ with the use of the assumption
EDIT: This does not solve your question as is pointed out in the comments
A: Here is my attempt. We have that
$$
\frac1n\sum_{k=1}^na_k=\frac1n\sum_{k=1}^{m-1}a_k+\frac1n\sum_{k=m}^na_k.
$$
By the C-S inequality,
$$
\frac1n\sum_{k=1}^{m-1}a_k
=\sum_{k=1}^{m-1}\frac{\sqrt k}n\frac{a_k}{\sqrt k}
\le
\Bigl[\frac1{n^2}\sum_{k=1}^{m-1}k\Bigr]^{1/2}
\Bigl[\sum_{k=1}^{m-1}\frac{a_k^2}{k}\Bigr]^{1/2}.
$$
The second term on the right hand side converges and the first term on the right hand side goes to $0$ as long as $m/n\to0$ as $n\to\infty$ since
$$
\frac1{n^2}\sum_{k=1}^{m-1}k=\frac{m(m-1)}{2n^2}.
$$
Similarly,
$$
\frac1n\sum_{k=m}^na_k=\sum_{k=m}^n\frac{\sqrt k}{n}\frac{a_k}{\sqrt k}
\le\Bigl[\frac{1}{n^2}\sum_{k=m}^nk\Bigr]^{1/2}\Bigl[\sum_{k=m}^n\frac{a_k^2}{k}\Bigr]^{1/2}.
$$
The first term converges to $1/2$ as long as $m/n\to0$ since
$$
\frac{1}{n^2}\sum_{k=m}^nk=\frac{(n-m+1)(n+m)}{2n^2}=\frac{n^2-m^2+m+n}{2n^2}
$$
and the second term converges to $0$ as $m\to\infty$. So we need to choose $m=m(n)$ such that $m(n)\to\infty$, but $m(n)/n\to0$  as $n\to\infty$.
I hope this is helpful.
