If the series $\sum_{k=1}^{\infty} a_k$ is Cesàro summable and $n a_n \to 0$ as $n \to \infty$, then the series converges I'm learning about Fourier series, specifically Cesàro summable sequences and series, and need help with the following problem: 

Show that if the series $\sum_{k=1}^{\infty} a_k$ is Cesàro summable and $n a_n \to 0$ as $n \to \infty$, then the series converges. 

The definition that is given for Cesàro summable series is as follows:
Cesàro summable series. A series $\sum_{k=1}^{\infty} a_k$ is called Cesàro summable if the sequence of partial sum $S_n$ is Cesàro convergent, i.e. if
$${S_0 + S_1 + \ldots + S_n \over n + 1}$$
converges as $n \to \infty$.
I'm sorry for the lack of effort on my part but I have no idea how to prove the above statement. 
 A: Yesterday I said that this was Tauber's Theorem, the original tauberian theorem. It's not; Tauber's Theorem is the analogous result for Abel summability. This is Someone's Theorem. I'm going to give a proof of ST organized in what I feel is the "right" manner, deriving convergence from a sort of "maximal inequality"; then as a bonus we will see that one can prove Tauber's Theorem in the same manner, using estimates that are a little more intricate, although still straightforward.
First, your notation seems a little curious, since you start the summation at $k=1$ but include $s_0$. I'm going to start at $k=1$ as you do, but instead consider the averages $$\sigma_n=\frac{s_1+\dots+s_n}{n}.$$
Someone's Theorem is immediate from the following:
Theorem If $na_n\to0$ then $s_n-\sigma_n\to0$.
For any sequence $a=(a_1,\dots)$ define $$Ma=\sup_n|na_n|.$$
The theorem follows from the following "maximal inequality":
Lemma $|s_n-\sigma_n|\le Ma$ for every $n$.
Proof: First note that $$\sigma_n=\sum_{k=1}^n\frac{n-k+1}{n}a_k.$$
(Simply count the number of times each $a_k$ appears in $\sigma_n$.) Hence
$$s_n-\sigma_n=\sum_{k=1}^n\frac{k-1}{n}a_k.$$Since $|k-1|\le k$ this shows that $$|s_n-\sigma_n|\le Ma\sum_{k=1}^n\frac1n=Ma.$$
QED.
Now to prove the theorem. Suppose $na_n\to0$. Let $\epsilon>0$. Choose $N$ so $|na_n|<\epsilon$ for all $n>N$, and define $$a_n'=\begin{cases}
a_n,&(1\le n\le N),
\\0,&(n>N)
\end{cases}$$and $$a_n''=a_n-a_n'.$$In what one hopes is transparent notation, it is clear that $$s_n'-\sigma_n'\to0$$and $$Ma''\le\epsilon.$$So
$$\limsup|s_n-\sigma_n|\le\limsup|s_n'-\sigma_n'|+\sup|s_n''-\sigma_n''|
\le Ma''\le\epsilon.$$Hence $\limsup|s_n-\sigma_n|=0$. QED.

Bonus: Tauber's Theorem
Say $$f(r)=\sum_{k=0}^\infty a_kr^k\quad(0<r<1)$$and $s_n=\sum_{k=0}^na_k$.
Tauber's Theorem If $na_n\to0$ and $\lim_{r\to1}f(r)=s$ then $\sum_{k=0}^\infty a_k=s$.
As with Someone's Theorem, this follows from the somewhat stronger result
Theorem If $na_n\to0$ then $s_n-f(1-1/n)\to0$.
And that follows by an argument as above if we can show that 
$$\left|s_n-f\left(1-\frac1n\right)\right|\le cMa.$$To begin, it's clear that $$\left|s_n-f\left(1-\frac1n\right)\right|\le Ma\sum_{k=1}^n\frac1k\left(1-\left(1-\frac1n\right)^k\right)
+Ma\sum_{k=n+1}^\infty\frac1k\left(1-\frac1n\right)^k,$$so we need only show that both sums on the right are bounded (independent of $n$).
There exist $\alpha$ and $\beta$ with $$0<\alpha<\left(1-\frac1n\right)^n<\beta<1\quad(n\ge2).$$So
$$
\sum_{k=1}^n\frac1k\left(1-\left(1-\frac1n\right)^k\right)\le
\sum_{k=1}^n\frac1k\left(1-\alpha^{k/n}\right)\le
\int_0^n\left(1-\alpha^{t/n}\right)\frac{dt}{t}
=\int_0^1\left(1-\alpha^{t}\right)\frac{dt}{t}$$
and similarly 
$$\sum_{k=n+1}^\infty\frac1k\left(1-\frac1n\right)^k
\le \sum_{k=n+1}^\infty\frac1k\beta^{k/n}
\le\int_n^\infty\beta^{t/n}\frac{dt}{t}
=\int_1^\infty\beta^t\frac{dt}{t}.$$
A: This is an exercise in Rudin's Principle of Mathematical Analysis (Exercise 3.14(d) page 80). 

If $\{s_n\}$ is a complex sequence, define its arithmetic means $\sigma_n$ by
  $$
\sigma_{n}=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1} \quad(n=0,1,2, \ldots)
$$
  Put $a_n = s_n-s_{n-1}$, for $n\ge 1$. Assume that $\lim (na_n)=0$ and that $\sigma_n$ converges. Then $\{s_n\}$ converges.

This is an easy consequence of Exercise 3.14(a), which asks the readers to show that standard convergence implies Cesaro convergence.
Proof. Observe that
$$
\begin{aligned} s_{n}-\sigma_{n} &=s_{n}-\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1} \\ &=\left(a_{0}+a_{1}+\cdots+a_{n-1}+a_{n}\right)-
 \frac{(n+1) a_{0}+n a_{1}+\cdots+2 a_{n-1}+a_{n}}{n+1} \\ 
&=\frac{a_{1}+2 a_{2}+\cdots+(n-1) a_{n-1}+n a_{n}}{n+1} \end{aligned}\;,
$$
which is a Cesaro mean of the sequence $\{na_n\}$. But it is assumed that $\lim (na_n)=0$. Thus, we have $\lim (\sigma_n-s_n)=0$ since standard convergence implies Cesaro convergence.

Note: One can derive the conclusion from a weaker hypothesis: Assume that the sequence $\{na_n\}$ is bounded. Then if $\lim \sigma_n = \sigma$, one has $\lim s_n = \sigma$. (This is the follow-up exercise 3.14(e) in Rudin.)
