Cesàro summability implies convergence of (conventional) sum when $\sum_{n=1}^\infty na_n^2<\infty$ 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?
thanks
 A: 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.
A: 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.
