Does $\sum\limits_{k=1}^{\infty} (a_{k+1} - a_k)^2 <\infty$ imply $\frac{1}{n^2} \sum\limits_{k=1}^n a_k^2 \to 0$? I'm currently working on some problem regarding Dirichlet forms and, as the title states, I'm trying to figure out if 
$$\sum\limits_{k=1}^{\infty} (a_{k+1} - a_k)^2 <\infty \Rightarrow \frac{1}{n^2} \sum\limits_{k=1}^n a_k^2 \to 0$$
holds. I've already tried it with some examples (mostly 'mainstream' sequences) and it seems to always work. However, I don't seem to be able to proof it and thus I tend to think the statement is false...
My initial idea for the proof was this:
We have $a_n = a_1 + \sum\limits_{k=1}^{n-1} a_{k+1} - a_k$.  Using $(a+b)^2 \leq 2(a^2+b^2)$ and the Cauchy-Schwarz inequality we get
$$a_n^2 = \left( a_1 + \sum\limits_{k=1}^{n-1} (a_{k+1} - a_k) \right)^2 \leq 2 a_1^2 + 2 \left( \sum\limits_{k=1}^{n-1} (a_{k+1}-a_k) \right)^2$$
$$\leq 2a_1^2 + 2 (n-1) \cdot \sum\limits_{k=1}^n (a_{k+1}-a_k)^2 \leq 2a^2_1 + C \cdot (n-1)$$
But this only implies that $\frac{1}{n^2} \sum\limits_{k=1}^n a_k^2$ is bounded. Any hints/ideas or a counterexample would be much appreciated.
 A: Here is a sledgehammer solution: Using the Hardy's inequality, we get
$$ \sum_{n=1}^{\infty} \frac{a_{n}^{2}}{n^{2}} < \infty. $$
Now you can apply the Kronecker's lemma to obtain the desired conclusion.
This at least shows that your guess is correct. I believe that there is a much simpler solution, and I will update my solution when I find it.
A: Here is a solution using Cauchy's inequality and Stolz–Cesàro theorem.
By Stolz–Cesàro theorem, we need to  prove 
$$\lim\dfrac{1}{n^2}\sum_{k=1}^n a_k^2 = \lim \dfrac{a^2_{n+1}}{2n+1} =0$$ provided the second limit exists.
Let $b_k = a_{k+1}- a_k$ with $b_0 = a_1$, we have $\sum_k b_k^2 < \infty$ and $a_{n+1} = \sum_{k=0}^{n}b_k$.
\begin{align}
a_{n+1}^2 = (\sum_{k=0}^{n}b_k)^2 \leq 2(\sum_{k=0}^{N}b_k)^2 + 2(\sum_{k=N+1}^{n}b_k)^2  \leq 2(\sum_{k=0}^{N}b_k)^2 + 2(n-N)(\sum_{k=N+1}^{n}b_k^2)
\end{align}
Thus we have
\begin{align}
\limsup_n\dfrac{a_{n+1}^2}{n} &\leq \limsup_n \dfrac{2(\sum_{k=0}^{N}b_k)^2 + 2(n-N)(\sum_{k=N+1}^{n}b_k^2)}{n}\\
&\leq 2(\sum_{k=N+1}^{\infty}b_k^2)
\end{align}
Since it's true for all $N$ and $\sum_{N}^\infty b_k^2 \to 0$ as $N\to \infty$, we have the conclusion.
