Let $\{a_n\}$ be a bounded and positive sequence. Show that

$$\lim_{n\to \infty}\frac{a_1+\cdots+a_n}{n}=0$$ if and only if $$\lim_{n\to \infty}\frac{a_1^2+\cdots+a_n^2}{n}=0.$$

My attempt:

The "$\Rightarrow$" is obvious. Note that $$\frac{a_1^2+\cdots+a_n^2}{n}\leq |M|\cdot\frac{a_1+\cdots+a_n}{n} $$ where $|M|$ is the bound of the sequence. So the convergence of the right side implies the convergence of the left side.

As for the converse direction, I really have no idea...

@kimchi lover points out using the Cauchy-Schwarz inequality and I had the following attempt...

$$\frac{a_1+\cdots+a_n}{n}=\frac{\frac{1}{\sqrt{n}}(a_1+\cdots+a_n)}{\frac{1}{\sqrt{n}}n}\leq \frac{(a_1^2+\cdots+a_n^2)(\frac{1}{n}+\cdots+\frac{1}{n})}{\sqrt{n}}$$

  • 6
    $\begingroup$ Cauchy-Schwarz ? $\endgroup$ Aug 14 '17 at 23:17

By Cauchy—Schwarz inequality, $$ \sum_{k=1}^n \frac{1}{n}\cdot a_k \leq \sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n \frac{1}{n^2}} = \sqrt{\sum_{k=1}^n a_k^2}\cdot\sqrt{\frac{1}{n}} = \sqrt{\frac{1}{n}\sum_{k=1}^n a_k^2} $$ and you can conclude by the squeeze theorem.

  • $\begingroup$ But doesn't that just show that if the RHS converges then the LHS converges? And the OP already showed that. We need for $\frac1n \sum a$ to be greater than something that we can relate to $\frac1n \sum a^2$ . $\endgroup$ Aug 14 '17 at 23:30
  • 2
    $\begingroup$ @MarkFischler No, the OP proved the other implication -- read their post again. $\endgroup$
    – Clement C.
    Aug 14 '17 at 23:32

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