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 $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=\lim\limits_{n\to \infty} x_n=0$.

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    $\begingroup$ The definition of the limit does not allow you to conclude that $\exists N \in \mathbb{N}$ such that $ n \geq N \implies |\frac{x_1^2+x_2^2+...+x_n^2}{n}|<\frac{1}{n^2}$. It does allow you to say that for each $m>0$, $\exists N \in \mathbb{N}$ such that $ n \geq N \implies |\frac{x_1^2+x_2^2+...+x_n^2}{n}|<\frac{1}{m^2}$, but this is a very different statement. $\endgroup$ – jawheele Apr 18 at 19:49

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

  • $\begingroup$ Thank you! I know that $\lim \limits_{n\to \infty} a_n=l <=> \forall \epsilon>0 \exists N$ such that $|a_n-l|<\epsilon$, $\forall n\ge N$. Could you tell me what is the reason why I can't choose $\epsilon=\frac{1}{n^2}$? $\endgroup$ – Math Guy Apr 18 at 20:03
  • $\begingroup$ @MathGuy: $N$ depends on $\epsilon$, but $\epsilon$ cannot vary with $n$. A simple example: $\frac 1n \to 0$,but $|\frac 1n| < \frac{1}{n^2}$ is never true. $\endgroup$ – Martin R Apr 18 at 20:06
  • $\begingroup$ Thank you for your clear explanation! $\endgroup$ – Math Guy Apr 18 at 20:16

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.

  • $\begingroup$ Thank you! I have also come up with this solution later on, but I am still curious why the other approach is wrong. $\endgroup$ – Math Guy Apr 18 at 20:05
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    $\begingroup$ Titu's lemma is Cauchy-Schwarz in disguise :) $\endgroup$ – Martin R Apr 18 at 20:08
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    $\begingroup$ @MartinR in fact, some books use Titu's lemma to prove Cauchy-Schwarz (it's a $\iff$). $\endgroup$ – rtybase Apr 18 at 20:09
  • $\begingroup$ @MathGuy I just provided a counterexample ... $\endgroup$ – rtybase Apr 18 at 20:36

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