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 $\lim\limits_{n\to \infty} \frac{x_1+x_2+…+x_n}{n}=\lim\limits_{n\to \infty} x_n=0$.
Sequence such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+…+x_n^2}{n}=0$
real-analysissequences-and-series
Best Answer
$\lim_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$ 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 $0 < p < q$.