Sequence such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$ Let $(x_n) _{n\ge 1}$ be a sequence of real numbers such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$. Prove that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=0$.
I used the definition of the limit to conclude that $\exists N\in \mathbb{N} $ such that $|\frac{x_1^2+x_2^2+...+x_n^2}{n}|<\frac{1}{n^2}$, $\forall n\ge N$. Hence, we get that $|x_1^2+x_2^2+...+x_n^2|<\frac{1}{n}$.
Now here comes the part where I am not really sure. I think that this implies that $\sum_{n=1}^{\infty}x_n^2=0$, and as a result $x_n\to 0$,which solves the problem because if we use the Stolz-Cesaro lemma we get that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=\lim\limits_{n\to \infty} x_n=0$.
 A: $\lim_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$
does not imply that $|\frac{x_1^2+x_2^2+...+x_n^2}{n}|<\frac{1}{n^2}$ for sufficiently large $n$. Also $\sum_{n=1}^{\infty}x_n^2=0$ would be true only if all $x_n$ are zero, so that approach cannot work, unfortunately.
But the Cauchy-Schwarz inequality gives
$$
\left |\sum_{k=1}^n 1\cdot x_k \right| \le \sqrt n \cdot \sqrt{\sum_{k=1}^n x_k^2}
$$
and therefore
$$
\left |\frac 1n \sum_{k=1}^n x_k \right| \le \sqrt{\frac 1n \sum_{k=1}^n x_k^2}
$$
This is also a special case of the Generalized mean inequality:
$$
\sqrt[p]{\frac 1n \sum_{k=1}^n x_k^p} \le \sqrt[q]{\frac 1n \sum_{k=1}^n x_k^q} 
$$
for non-negative  real numbers $x_1, \ldots, x_n$ and $0 < p < q$.
A: Alternatively, you can use Titu's lemma (and here)
$$\frac{x_1^2+x_2^2+...+x_n^2}{n}= 
\frac{1}{n}\left(\sum\limits_{k=1}^n\frac{x_k^2}{1}\right)\ge
\frac{1}{n}\frac{\left(\sum\limits_{k=1}^nx_k\right)^2}{\sum\limits_{k=1}^n1}=\\
\left(\frac{x_1+x_2+...+x_n}{n}\right)^2$$

To the question of why $$\lim\limits_{n\rightarrow\infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0 \ \nRightarrow \ \left|\frac{x_1^2+x_2^2+...+x_n^2}{n}\right|<\frac{1}{n^2}$$
Consider the case $x_n=\frac{1}{\sqrt{n}}$. Then $x_1^2+x_2^2+...+x_n^2=1+\frac{1}{2}+...+\frac{1}{n}$ which is the harmonic series, with
$$\ln{n}+1>1+\frac{1}{2}+...+\frac{1}{n}>\ln{(n+1)} \Rightarrow \lim\limits_{n\rightarrow\infty}\frac{1+\frac{1}{2}+...+\frac{1}{n}}{n}=0$$
but this also leads to
$$\frac{\ln{(n+1)}}{n}< \left|\frac{x_1^2+x_2^2+...+x_n^2}{n}\right|<\frac{1}{n^2}$$
which is wrong.
