Why does the following Fourier series does not converge for $x \in R$, and does for $x \in [0,2\pi]$? I would really love your help with the following facts that I can't understand.
I can't understand why the following Fourier series does not converge:
$$\sum_{0}^{\infty}\frac{e^{inx}}{n^2}.$$
1.If I use the fact that $e^{inx}= \cos nx + i \sin nx$ so I get that the sum equals the following sum of sums: $\sum_{0}^{\infty} \frac{\cos nx}{n^2}+\sum_{0}^{\infty} \frac{\sin nx}{n^2}$, and here: why each of the sums does not converge? Can't I use Dirichlet test? $\frac{1}{n^2}\to0 , \sum_{0}^{N}|\sin nx|<M$
2.Why does it converge for $x\in [0,2\pi]$?
Thanks a lot. 
 A: If you leave out the $n=0$ term, then the series converges absolutely for all $x\in\mathbb{R}$ since
$$
\left|\sum_{n=1}^\infty\frac{e^{inx}}{n^2}\right|\le\sum_{n=1}^\infty\left|\frac{e^{inx}}{n^2}\right|=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}
$$
A: The series
$$
\sum_{n=1}^\infty \frac{e^{inx}}{n^2}
$$
converges absolutely if $x$ is real, and hence if $0\le x\le2\pi$.  (But in order to say that, I had to start with $n=1$; I'm guessing the $0$ in the question is a typo and $1$ was intended.)  Absolute convergence of this series means
$$
\sum_{n=1}^\infty \left|\frac{e^{inx}}{n^2}\right| < \infty.
$$
That is true because $\left|\dfrac{e^{inx}}{n^2}\right|= \dfrac{1}{n^2}$, and $\sum\limits_{n=1}^\infty \dfrac1{n^2} <\infty$.  A theorem in mathematical analysis says that if a series converges absolutely, then it converges.
However, if $x$ is not real, then the series may diverge.  For example, if $x=-i\log_e 2$, then the series becomes $\sum\limits_{n=1}^\infty \dfrac{2^n}{n^2}$.  The ratio test shows that that diverges.
