If $\sum a_k^2 /k$ converges, then $1/N \sum_1^{N}a_k \to 0$ I want to show that if $\sum a_k^2 / k$ converges, then $1/N \sum_{1}^Na_k \to 0$.
Now, if $a_n \to 0$, then the result follows. But of course $a_n\to 0$ is not a necessary condition for $\sum a_{n}^2/n$ to converge. We might want to use Cauchy-Schwarz to say that $a_k^2/k \leq a_k^4 + 1/k^2$, but of course the inequality is pointing the wrong way. Any ideas?
 A: By your condition, for any $\epsilon>0$, there exists $m$ such that when $N\geq m$ 
$\dfrac{a_{m+1}^2}{m+1}+...+\dfrac{a_{N}^2}{N}<\epsilon $
By Cauchy one has
$(a_{m+1}+...a_{N})^2<(\dfrac{a_{m+1}^2}{m+1}+...+\dfrac{a_{N}^2}{N})(m+1+m+2+...+N)$.
so,
$(a_{m+1}+...+a_{N})^2<\epsilon \cdot \dfrac{(m+1+N)(N-m)}{2}<N^2 \cdot \epsilon \cdot constant$.
When $N$ is big enough compared with $m$,
$\dfrac{(a_{m+1}+...+a_{N})^2}{N^2}<\epsilon \cdot constant$
Therefore the limit in question is zero.
A: Split the sum at $k = \sqrt{N}$. 
If $\sqrt{N} < k < N$ then by Cauchy-Schwarz
$$
\sum_{\sqrt{N} < k \leq N} a_k \leq N^{1/2} \cdot \bigg ( \sum_{\sqrt{N} < k \leq N}
a_k^2 \bigg )^{1/2}
$$
And notice that
$$
\sum_{\sqrt{N} < k \leq N} a_k^2 \leq N \sum_{N > k > \sqrt{N}} \frac{a_k^2}{k}
= o(N)
$$
because $\sum_{k > \sqrt{N}} a_k^2 / k = o(1)$. 
If $k < \sqrt{N}$ then
$$
\sum_{k \leq \sqrt{N}} a_k \leq \bigg ( \sum_{k \leq \sqrt{N}} k \bigg )^{1/2}
\cdot \bigg ( \sum_{k} \frac{a_k^2}{k} \bigg )^{1/2} \leq C N^{1/2}
$$
with $C = (\sum a_k^2 / k)^{1/2}$.
Combining the two sums we conclude that
$$
\sum_{k \leq N} a_k \leq C N^{1/2} + o(N) = o(N)
$$
This is the desired claim. Note that we could have splitted at any $k = f(N)$ with $f$ going to infinity.
