Last 2016 issue, are you able to solve it?

We prove a more general statement: If p is a positive integer and $(x_{n})$ , $n\leq 1$ is a sequence such that $\;$
$\lim_{n \to \infty } \frac{x _{1}^{2p}+...+x_{n}^{2p}}{n}=0$ $(1)$
$\;$ Then
$\lim_{n \to \infty} \frac{x _{1}+...+x_{n}}{n}=0$ $(2)$
For this, recall the inequality
$\;$
$(\frac{x_{1}+\cdots+x_{n}}{n})^{2p}$ $\leq$ $\frac {x_{1}^{2p}+...+x_{n}^{2p}}{n}$
$\;$
It follows that
$\;$
$\left | \frac{x_{1}+\cdots+x_{n}}{n} \right |\leq \sqrt[2p]{\frac{x_{1}^{2p}+...+x_{2}^{2p}}{n}}$
$\;$
Using the squeeze theorem and the hypothesis $(1)$ , the conclusion $(2)$ follows. For p=1 we obtain the intial problem.
$\;$
The converse is not true. Take $x_{n}=(-1)^{n}$ and observe that:
$\;$
$\frac{x_{1}+\cdots+x_{n}}{n}=0$ if n is even and $\frac{-1}{n}$ if n is odd.
$\;$
Hence
$\lim_{n \to\infty} \frac{x _{1}+\infty+x_{n}}{n}=0$
$\;$
but
$\frac{x _{1}^{2}+\cdots +x_{n}^{2} } {n}=1$

The given fact: $\lim_{n\to \infty}\dfrac{\displaystyle \sum_{k=1}^{n}x^{2}_{k}}{n}$ As this limit converges, this means, that the series $\sum_{n=1}^{\infty}x^{2}_{n}$ converges, but this doesn't mean $\sum_{n=1}^{\infty}x_{n}$ converges. (Example: $x_{n}=\frac{1}{n}$ ) For the limit: $\lim_{n\to \infty} \dfrac{\displaystyle \sum_{k=1}^{n}x_{n}}{n}$ To exist, we need $\sum_{n=1}^{\infty}x_{n}$ to converge, but we don't have additional information, so the limit above, doesn't necessary exist.