For a given sequence $(a_k)$, there is no Riemann integrable function f such that $\hat{f}(k) = a_k \forall k$ I'm working out of Stein's Fourier Analysis: An Introduction, and am on chapter 3.  There is an exercise that gives us a specific sequence $(a_k)$ and asks us to show that $\sum\nolimits_{k=-\infty}^{\infty} |a_k|^2$ converges but that no Riemann integrable function has $k$th Fourier coefficient equal to $a_k \forall k$
My question is, how do I show the second part?  It isn't clear from the text what necessary conditions are on Fourier coefficients, but maybe I'm missing something.
Edit - The sequence in question is
$a_k = \begin{cases} 1/k, & k \geq 1 \\ 0, & k \leq 0 \end{cases}$
$(a_k) \in l^2(\mathbb{Z})$ since the sum $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges, but I'm not sure about the second part.
If $(a_k)$ was the Fourier coefficients for $f$, then $f(x) - S_n(f)(x)$ converges to $0$ in the $L^2$ norm where $S_n = \sum_{k=-n}^{n}a_ke^{ikx} = \sum_{k=1}^{n}f(x)e^{ikx}/k$  
Should I show that $f-S_n$ does not converge to 0 in the $L^2$ norm? Or is there some other condition I should be checking?
 A: Besides being square-summable, there are no  easy necessary conditions for $a_k$ to be the Fourier coefficients of a Riemann integrable function. 

Should I show that $f−S_n$ does not converge to $0$ in the $L^2$ norm?

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.
