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

• 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. – 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 $$0 < p < q$$.

• 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 $\epsilon=\frac{1}{n^2}$? – Math Guy Apr 18 at 20:03
• @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. – Martin R Apr 18 at 20:06
• Thank you for your clear explanation! – 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.

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