Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that

If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero.

A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.

Could anyone please help me to prove it?

  • $\begingroup$ Note that if $\sum_1^\infty a_n^2/n$ converges, then every partial sum converges. $\endgroup$ – Kumar Jul 10 '19 at 14:27
  • $\begingroup$ I think you should go through the answers of this question. It will help you to generate ideas: math.stackexchange.com/questions/4603/… $\endgroup$ – Kumar Jul 10 '19 at 14:46
  • 1
    $\begingroup$ @Kumar the link deals with nonincreasing sequences....There is no way to apply the contents to arbitrary sequences.... $\endgroup$ – Keith Jul 10 '19 at 15:10

Fix any $N_0$ and write: \begin{align*} \frac{1}{N} \sum\limits_{n = 1}^N |a_n| &= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N |a_n| }{N} \\ &= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N \frac{|a_n|}{\sqrt{n}} \cdot \sqrt{n} }{N} \\ &\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n} \right) \left(\sum\limits_{n = N_0 + 1}^N n \right) }}{N} \\ &\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n} \right)}. \end{align*} Here on the third line I use CS inequality. To obtain the forth line I simply bound $\sum\limits_{n = N_0 + 1}^N n$ from above by $N^2$.

Now by taking $N\to\infty$ we get $$\limsup\limits_{N\to\infty} \frac{1}{N} \sum\limits_{n = 1}^N |a_n| \le \sqrt{\left(\sum\limits_{n = N_0 + 1}^\infty \frac{a_n^2}{n} \right)}. $$ As $N_0$ can chosen arbitrarily, the right hand side can be made arbitrarily small --- here we use the fact that $\sum\limits_{n = 1}^\infty \frac{a_n^2}{n} $ is convergent.

  • $\begingroup$ Oh....It is very ingenious... $\endgroup$ – Keith Jul 10 '19 at 15:18
  • $\begingroup$ Splitting and $\limsup$, nice approach. $\endgroup$ – Gabriel Romon Jul 10 '19 at 15:24

We use the Cauchy-Schwartz inequality to write

$$ \frac{1}{N}\sum_{n=1}^{N}a_n = \sum_{n=1}^{N} \frac{\sqrt{n}}{N}\frac{a_n}{\sqrt{n}} \leq \sqrt{\sum_{n=1}^N \frac{n}{N^2} \sum_{n=1}^{N} \frac{a^2_n}{n} } = \sqrt{\frac{1}{N^2}\frac{N(N+1)}{2}\sum_{n=1}^{N}\frac{a_n^2}{n}}$$

Which converges as $N \to \infty$ with the use of the assumption

EDIT: This does not solve your question as is pointed out in the comments

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    $\begingroup$ But at best this tells you that $$\lim_{N\to\infty} \frac 1 N \sum^N_{n=1} a_n \le \frac 12 \sum^\infty_{n=1} \frac{a_n^2}{n}.$$ The question is asking us to prove that $$\lim_{N\to\infty} \frac 1 N \sum^N_{n=1} a_n = 0.$$ $\endgroup$ – User8128 Jul 10 '19 at 14:35

Here is my attempt. We have that $$ \frac1n\sum_{k=1}^na_k=\frac1n\sum_{k=1}^{m-1}a_k+\frac1n\sum_{k=m}^na_k. $$ By the C-S inequality, $$ \frac1n\sum_{k=1}^{m-1}a_k =\sum_{k=1}^{m-1}\frac{\sqrt k}n\frac{a_k}{\sqrt k} \le \Bigl[\frac1{n^2}\sum_{k=1}^{m-1}k\Bigr]^{1/2} \Bigl[\sum_{k=1}^{m-1}\frac{a_k^2}{k}\Bigr]^{1/2}. $$ The second term on the right hand side converges and the first term on the right hand side goes to $0$ as long as $m/n\to0$ as $n\to\infty$ since $$ \frac1{n^2}\sum_{k=1}^{m-1}k=\frac{m(m-1)}{2n^2}. $$ Similarly, $$ \frac1n\sum_{k=m}^na_k=\sum_{k=m}^n\frac{\sqrt k}{n}\frac{a_k}{\sqrt k} \le\Bigl[\frac{1}{n^2}\sum_{k=m}^nk\Bigr]^{1/2}\Bigl[\sum_{k=m}^n\frac{a_k^2}{k}\Bigr]^{1/2}. $$ The first term converges to $1/2$ as long as $m/n\to0$ since $$ \frac{1}{n^2}\sum_{k=m}^nk=\frac{(n-m+1)(n+m)}{2n^2}=\frac{n^2-m^2+m+n}{2n^2} $$ and the second term converges to $0$ as $m\to\infty$. So we need to choose $m=m(n)$ such that $m(n)\to\infty$, but $m(n)/n\to0$ as $n\to\infty$.

I hope this is helpful.


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