Series which are not Fourier Series How to show that
$$
\sum_{n=2}^\infty \frac{\sin{(nx)}}{\log n}
$$
not the Fourier series of any function?

I have shown that the series is convergent by Dirichlet test.

Let $a(n)=\frac{1}{\log n}$.
What is $\sum (a(n))^2$, to apply Parseval's theorem?
Since $\Bigl\lvert\sum\sin(n x)\Bigr\rvert\leq M$ for $M&lt;\infty$ and $\Bigl\{\frac{1}{\log(n)}\Bigr\}\to 0$ monotonically, by Dirichlet's test, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ converges.



If $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is a Fourier series, by Parseval's theorem, there exist a Riemann integrable function $f$ such that 
$$
\int_{-\pi}^{\pi}\lvert f(x)\rvert^2dx=2\pi\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert}
$$

Theorem:

Suppose $a_1\geq a_2\geq\cdots\geq 0$. Then the series $\sum a_n$ converges iff $\sum 2^ka_{2^k}$ converges.
$$
\sum_{n=2}^{\infty}\frac{2^k}{k^2\log^2(2)}\geq\sum_{n=2}^{\infty}\frac{1}{k}=\infty
$$



Therefore, by Cauchy's condensation test, the $\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert}$ does not converge which contradicts Parseval's theorem. Thus, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is not a Fourier series.
Suppose that it is the Fourier series of $f(x)$ in $(-\pi,\pi)$, namely,
$$ f(x)=\sum_{n=2}^\infty \frac{\sin(nx)}{\log n},\quad x\in(-\pi,\pi).$$
Then by Pareraval's Identity,
$$\int_{-\pi}^\pi f(x) \, dx=\sum_{n=2}^\infty\frac{1}{\log^2n}=\infty.$$