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

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.

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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 $$\sum_{k=n+1}^\infty\frac1k\left(1-\frac1n\right)^k \le \sum_{k=n+1}^\infty\frac1k\beta^{k/n} \le\int_n^\infty\beta^{t/n}\frac{dt}{t} =\int_1^\infty\beta^t\frac{dt}{t}.$$