# Laurent Series and finding values of the specified sum.

Comparing the coefficients in the Laurent developments of $cot(\pi z)$ and its expression as a sum of partial fractions, find the values of $\sum_{n=1}^\infty$ $\frac{1}{n^{4}}$ and $\sum_{n=1}^\infty$ $\frac{1}{n^{6}}$. I am struggling with this problem. I do not know where to start. Any help would be wonderful.

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see this thread –  Raymond Manzoni Mar 6 '12 at 1:18
I'm afraid I didn't quite follow the thread. Which part am I supposed to be following? –  Shalya Mar 6 '12 at 1:31

Computing the Fourier series of $\cos(zx)$ for $-\pi \le x \le \pi$ returns :

$$\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}+\frac{\cos(1x)}{1^2-z^2}-\frac{\cos(2x)}{2^2-z^2}+\frac{\cos(3x)}{3^2-z^2}-\cdots\right]$$ (details for the Fourier computation here)

applying this to $x=\pi$ gives : $$\cot(\pi z)=\frac1{\pi}\left[\frac1{z}-\sum_{k=1}^{\infty}\frac{2z}{k^2-z^2}\right]$$ that we may rewrite as : $$\pi z\cot(\pi z)=1-2\sum_{k=1}^{\infty}\frac{z^2}{k^2-z^2}$$ expand $\frac{z^2}{k^2-z^2}$ in Taylor series (for $z < k$) : $\displaystyle \frac{z^2}{k^2-z^2}=\sum_{i=1}^\infty \frac{z^{2i}}{k^{2i}}$ so that :

$$\pi z\cot(\pi z)=1-2\sum_{k=1}^{\infty}\sum_{i=1}^\infty \frac{z^{2i}}{k^{2i}}=1-2\sum_{i=1}^{\infty}\left(\sum_{k=1}^\infty \frac 1{k^{2i}}\right)z^{2i}$$ $$\frac{\pi z}2 \cot(\pi z)= \frac 12-\sum_{i=1}^{\infty}\zeta(2i)z^{2i}$$

At this point you'll just have to expand $\frac{\pi z}2 \cot(\pi z)$ in powers of $z$ and identify the coefficients of both expansions (for the expansion of $\cot(x)$ search 'cotangent' at Wikipedia or deduce it from $\cot(x)=\tan\left(\frac{\pi}2-x\right)$ or from $\cot'(x)=-1-\cot(x)^2$).

(this was more or less my answer combined with the start of the answer of robjohn in the other thread).

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