If the average of a positive sequence converges to zero, does the average of the square converge to zero?

Question

Give $\frac{\sum\limits_{i=1}^n x_i}{n}\rightarrow 0$, does it imply that $\frac{\sum\limits_{i=1}^n x_i^2}{n}\rightarrow 0?$ if $x_i\geq 0$.

Thank you @D.B. point out a counter example when $x_i = (-1)^i$. If we are given additional conditions $x_i\geq 0$, does the convergence hold?

Answer 1

The question is whether there is a sequence $(x_k)$ of nonnegative real numbers such that, as $n$ approaches infinity, $$\frac{\sum\limits_{k=1}^n x_k}{n}$$ approaches zero, but $$\frac{\sum\limits_{k=1}^n x_k^2}{n}$$ does not approach zero.

The answer is "yes".$\;$Here is an example . . .

For each positive integer $k$, let $$ x_k= \begin{cases} \sqrt[3]{k}&\text{if$\;k\;$is a perfect cube}\\ 0&\text{otherwise} \end{cases} $$ and for each positive integer $n$, let $c_n=\left\lfloor{\sqrt[3]{n}}\right\rfloor$. \begin{align*} \text{Then}\;\;\frac{\sum_{k=1}^n x_k}{n} &=\frac{\sum_{k=1}^{c_n} c_k}{n} \qquad\qquad\;\;\;\;\; \\[4pt] &\le \frac {\sum_{k=1}^{c_n} c_k} {c_n^3}\\[4pt] &=\frac{c_n(c_n+1)}{2c_n^3}\\[4pt] \end{align*} which approaches zero, as $n$ approaches infinity. \begin{align*} \text{But}\;\;\frac{\sum_{k=1}^n x_k^2}{n} &=\frac{\sum_{k=1}^{c_n} c_k^2}{n}\\[4pt] &> \frac {\sum_{k=1}^{c_n} c_k^2} {(c_n+1)^3}\\[4pt] &\ge \frac {\sum_{k=1}^{c_n} c_k^2} {8c_n^3}\\[4pt] &= \frac {c_n(c_n+1)(2c_n+1)} {48c_n^3}\\[4pt] \end{align*} which is bounded below by ${\large{\frac{1}{24}}}$, hence the $\text{LHS}$ does not approach zero, as $n$ approaches infinity.

Answer 2

No. Consider the sequence where the $x_i = (-1)^i$. Then, the first sequence converges, but the second diverges.

Answer 3

If you want the series themselves to converge (without the division by $n$), this means the $x_i$ must eventually become smaller than $1$ and thus $x_j^2<x_j$ for all $j>N \in \mathbb{N}$. In that case your conjecture would hold true.