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
$$\sum_{k=n+1}^\infty\frac1k\left(1-\frac1n\right)^k
\le \sum_{k=n+1}^\infty\frac1k\beta^{k/n}
\le\int_n^\infty\beta^{t/n}\frac{dt}{t}
=\int_1^\infty\beta^t\frac{dt}{t}.$$
Best Answer
Besides being square-summable, there are no easy necessary conditions for $a_k$ to be the Fourier coefficients of a Riemann integrable function.
No, because it does converge. Every square-summable sequence of coefficients is the Fourier series of some $L^2$ function $f$, to which the partial sums converge in the $L^2$ norm. The thing is, our $f$ is not Riemann integrable because it's not bounded.
Think of the domain of $f$ as the unit circle on the complex plane. The series in question is $\sum_{k=1}^\infty k^{-1}e^{ik\theta}$, which we can write as $\sum_{k=1}^\infty k^{-1}z^k$, with $z=e^{i\theta}$. The latter series converges in the open unit disk and defines a holomorphic function $F$ there. We can find $F$ explicitly as follows: $$F'(z) = \sum_{k=1}^\infty z^{k-1} = \frac{1}{1-z}$$ $$F(z) = \int_0^z F'(\zeta)\,d\zeta = -\log (1-z)$$ The boundary values of $F$ are not bounded: $$ \operatorname{Re}\log (1-e^{i\theta}) =\frac12 \log|1-e^{i\theta}|^2 = \frac12 \log(2-2\cos \theta)$$
Of course, some justification is needed for the claim that the boundary values of $F$ (taken as radial limits) recover the function $f$ that we started with. I don't know what the author of the book wanted the readers to do here, because I don't have the book.