Sequence such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$

Solution 1:

$\lim_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$ 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 $0 < p < q$.

Solution 2:

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.