# 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.

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Why do you think it doesn't converge? There's obviously a problem of definition since you have $n=0$ as one of the terms – Thomas Andrews Dec 28 '11 at 17:51
Both real and imaginary part converge indeed (if the sum starts at n=1 at least) respectively to $(\pi-x)^2/4-\pi^2/12$ and $Cl_2(x)= -\int^x_0 \log(2*\sin(t/2))\ dt$ (the Clausen integral (mathworld.wolfram.com/ClausensIntegral.html)) – Raymond Manzoni Dec 28 '11 at 18:15
@Jozef : If you put the period at the end of the sentence OUTSIDE of the "displayed" $\TeX$, then it ends up at the beginning of the next line, and the rest of that line will be blank. I fixed that. – Michael Hardy Dec 28 '11 at 18:57
Thanks all. you were very helpful. – Jozef Dec 29 '11 at 6:35

## 2 Answers

If you leave out the $n=0$ term, then the series converges absolutely for all $x\in\mathbb{R}$ since $$\left|\sum_{n=1}^\infty\frac{e^{inx}}{n^2}\right|\le\sum_{n=1}^\infty\left|\frac{e^{inx}}{n^2}\right|=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

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Thanks a lot rob. – Jozef Dec 29 '11 at 6:35

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.

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Thank you. very helpful. – Jozef Dec 29 '11 at 6:35