Suppose that series $\sum_{n=1}^\infty na_n^2<\infty$ and $\sum_{n=1}^\infty a_n$ is Cesàro summable. How do you show that $\sum_{n=1}^\infty a_n$ converges? I know that if $\lim_{n\rightarrow \infty} na_n=0$ then the result is true, is this any useful in here?



Write $S_N = \sum_{n=1}^N a_n$. Then, $$ \left| S_N - \frac{1}{N} \sum_{n=1}^N S_n \right|^2 = \left|\sum_{n=1}^N \frac{n-1}{N} a_n \right|^2.$$ This is less than $$\sum_{n=1}^N \frac{(n-1)^2}{N} |a_n|^2$$ using Cauchy's inequality, which is then bounded by $\sum_{n=1}^N n |a_n|^2$, giving the convergence.

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    $\begingroup$ Well I've been quite blind… Thank you and +bounty! $\endgroup$ – user42268 Mar 22 '16 at 6:44
  • $\begingroup$ Could you elaborate on the final step in your proof? It looks to me as though you're claiming that obtaining an upper bound on $\left|S_N-\frac1N\sum_{n=1}^N S_n\right|$ which doesn't depend on $N$ is enough to prove $\{S_N\}$ converges. This is false: Take $a_n=(-1)^n$, so that $S_N$ is zero if $N$ is even and $-1$ if $N$ is odd. Then $\left|S_N-\frac1N\sum_{n=1}^N S_n\right|\le\frac12$ for all $N$, but $\{S_N\}$ fails to converge. $\endgroup$ – A. Howells Aug 9 '18 at 23:48
  • $\begingroup$ @A.Howells You're right, it only shows boundedness. I'm not sure exactly how to complete the argument but my guess would be that if $S_N$ is bounded but not convergent, then $a_n$ has to be somewhat alternating which will contradict $\sum na_n^2 < \infty$. The problem here is to make precise this "somewhat alternating". $\endgroup$ – HEXQ Aug 10 '18 at 6:47
  • $\begingroup$ I know of two methods to show that your last out-of-line equation approaches 0 as $N\to\infty$. The first is to use the lemma in my answer (below). The second can be found in the middle of this answer. $\endgroup$ – A. Howells Aug 10 '18 at 14:18

Here is a complete solution. The missing piece was inspired by this answer from user mvggz. To begin, we prove:

Lemma: Suppose that $\{b_n\}_{n\in\mathbb N}$ is a sequence of non-negative real numbers satisfying $\sum_{n=1}^\infty b_n<\infty$. Then $$\lim_{N\to\infty}N^{-1}\sum_{n=1}^N n b_n=0.$$

Proof: Fix $\varepsilon>0$. Using convergence of $\sum_{n=1}^\infty b_n$, pick $m\in\mathbb N$ large so that $\sum_{n=m}^\infty b_n<\varepsilon/2$. Now pick $M\in\mathbb N$ larger than $m$ such that $$M>2\varepsilon^{-1}\sum_{n=1}^{m-1}nb_n.$$ Observe that for all $N\ge M$, we have $$\begin{align} N^{-1}\sum_{n=1}^N n b_n &=N^{-1}\sum_{n=1}^{m-1} n b_n+\sum_{n=m}^N\frac nN b_n\\ &\le M^{-1}\sum_{n=1}^{m-1} n b_n+\sum_{n=m}^N b_n\\ &<\frac\varepsilon2+\sum_{n=m}^\infty b_n\\ &<\varepsilon, \end{align}$$ as desired.

Proposition: Suppose that $\{a_n\}_{n\in\mathbb N}$ is a sequence of complex numbers such that $\sum_{n=1}^\infty a_n$ is Cesàro summable to $A\in\mathbb C$, and such that $\sum_{n=1}^\infty n|a_n|^2<\infty$. Then $\sum_{n=1}^\infty a_n=A$.

Proof: Consider the square of the difference between the $N$-th partial sum of $a_n$ and the $N$-th Cesàro average. Specifically, observe that by Cauchy-Schwarz $$\begin{align} \left|\sum_{n=1}^N a_n - \frac1N\sum_{n=1}^N \sum_{k=1}^n a_k \right|^2 &=\left|\frac1N\sum_{n=1}^N Na_n - \frac1N\sum_{n=1}^N (N-n+1)a_n \right|^2\\ &=\left|\sum_{n=1}^N \frac{n-1}N a_n \right|^2\\ &\le\left(\sum_{n=1}^N1^2\right)\left(\sum_{n=1}^N \frac{(n-1)^2}{N^2} |a_n|^2 \right)\\ &=\sum_{n=1}^N \frac{(n-1)^2}N |a_n|^2\\ &\le\sum_{n=1}^N \frac{n^2}N |a_n|^2; \end{align}$$ this approaches zero as $N\to\infty$ by our lemma (take $b_n=n|a_n|^2$). Since $$ \lim_{N\to\infty}\frac1N\sum_{n=1}^N \sum_{k=1}^n a_k=A $$ by assumption, we conclude that $\sum_{n=1}^\infty a_n=A$, as desired.


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