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let $f$ be a $2 \pi$ periodic function with $$ f(t) = \begin{cases} 1, & 0 \leq |t| \leq \dfrac{2 \pi} 3, \\ 0, & \text{otherwise}. \end{cases} $$

Compute the fourier series of this function,

so based on my calculation, the fourier coefficient I got is

$$\hat{f}(n) = \frac 1 {\pi n} \sin\left( \frac{2n\pi} 3 \right)$$

The fourier series will be

$$\sum_{n \geq 0} \frac 1 {\pi n} \sin\left( \frac{2n \pi} 3 \right)(e^{in \theta} - e^{-in \theta})$$

so I'm just wondering, did I do something wrong here because something looks off.

Because then I need to evaluate $$\sum \limits_{n \geq 0} \dfrac{1}{(3n+1)(3n+2)}$$ but I don't see how to get there

Nevermind I figured it out

I did the series wrong, The actual one is

$$\sum_{n \geq 0} \frac 1 {\pi n} \sin\left( \frac{2n \pi} 3 \right)(e^{in \theta} + e^{-in \theta}) + \frac{2}{3}$$

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  • $\begingroup$ If you solved the question by yourself maybe it would be a good idea to answer it by yourself here. That way, it won't appear as unanswered. $\endgroup$ – NeedForHelp Feb 22 '17 at 3:13
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Your Fourier coefficients are OK: $$ \hat{f}(n)=\begin{cases}\frac{1}{n \pi}\sin(\frac{2\pi n}{3}) & n\neq0 \\ \frac{2}{3} & n=0\end{cases} $$ Since $f$ is differentiable at $0$, we know its Fourier series at $0$ converges to $f(0)$. That is, \begin{align*} 1 &= \frac{2}{3} + \frac{2}{\pi}\sum_{n\geq0}\left(\frac{1}{3n+1}\frac{\sqrt{3}}{2}-\frac{1}{3n+2}\frac{\sqrt{3}}{2}\right)\\ &=\frac{2}{3}+\frac{\sqrt{3}}{\pi}\sum_{n\geq0}\frac{1}{(3n+1)(3n+2)} \end{align*} hence $$ \sum_{n\geq0}\frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$

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