Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.