Prove that $\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0$ Let's $x_n\ge0$ and $$\overline{\lim}_{n\rightarrow\infty}\frac{x_1+x_2+\cdots+x_n}{n}\lt+\infty,~~\lim_{n\rightarrow\infty}\frac{x_n}{n}=0.$$Prove that $$\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0.$$
I have no idea how I can start the proof.
Thanks!
 A: Your first requirement implies that there is some $R\in\left(0,\infty\right)$
such that
$$
\sum_{i=1}^{n}\frac{x_{i}}{n}\leq R
$$
holds for all $n\in\mathbb{N}$.
Now, let $\varepsilon>0$ be arbitrary. Your second requirement shows that there is some $n_{1}\in\mathbb{N}$ such that $\frac{x_{n}}{n}<\frac{\varepsilon}{2R}$ holds for all $n\geq n_{1}$.

 Let $S:=\max_{i=1,\dots,n_{1}}x_{i}$ and choose $n_{2}\in\mathbb{N}$ with $n_{2}\geq n_{1}$ and $\frac{SR}{n_{2}}\leq\varepsilon/2$.

We now derive for $n>n_{2}\geq n_{1}$ that
\begin{eqnarray*}
\sum_{i=1}^{n}\frac{x_{i}^{2}}{n^{2}} & = & \left[\max_{i=1,\dots,n_{1}}x_{i}\right]\cdot\frac{1}{n}\cdot\sum_{i=1}^{n_{1}}\frac{x_{i}}{n}+\left[\max_{j=n_{1}+1,\dots,n}\frac{x_{j}}{n}\right]\cdot\sum_{i=n_{1}+1}^{n}\frac{x_{i}}{n}.
\end{eqnarray*}
I believe you can take it from there. Otherwise:

\begin{eqnarray*}\dots \\ & \leq & \frac{SR}{n}+\left[\max_{j=n_{1}+1,\dots,n}\frac{x_{j}}{j}\right]\cdot\sum_{i=1}^{n}\frac{x_{i}}{n}\\ & \leq & \frac{SR}{n}+\frac{\varepsilon}{2R}\cdot R\leq\varepsilon.\end{eqnarray*}

Remark: What we are doing in the above is essentially an interpolation argument between an $\ell^1$ estimate and an $\ell^\infty$ estimate to get an $\ell^2$ estimate . The only problem is that the $\ell^\infty$ estimate only holds in the limit, so that one has to treat the first finitely many terms in a special way.
A: Another approach: use the weighted variant of Cauchy's limit theorem (see page 34 of Knopp's book)

*

*Set $y_n=x_n+1$, then  $y_n$ has the same properties as $x_n$ in the original question.

*$z_n=\frac{y_1\frac{y_1}{1}+\ldots y_n \frac{y_n}{n}}{y_1+\ldots +y_n} \to 0$  (by Cauchy's theorem)

*$u_n=\frac{\frac{y_1^2+\ldots y_n^2}{n}} {y_1+\ldots +y_n} \le z_n$ (a trivial lower bound for the numerator)

*$\frac{\frac{y_1^2+\ldots y_n^2}{n}} {Kn} < u_n$ (since $y_1+\ldots+y_n<Kn$ for some $K>0$)

*By noting that $y_n^2$ majorizes $x_n^2$ we can deduce the needed result.

