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I am self-studying complex analysis, so I am a rookie. I ran across an interesting series I am trying to evaluate using CA.

Show that $$\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{7}}=\frac{19{\pi}^{7}}{56700}$$

I began by considering $$\oint_{C_{N}}\frac{\pi \cot(\pi z)\coth(\pi z)}{z^{7}}$$

$$=\oint_{C_{N}}\frac{\pi \cos(\pi z)\cosh(\pi z)}{z^{7}\sin(\pi z)\sinh(\pi z)}$$

Where $C_{N}$ is the square centered at the origin with vertices

$$(N+1/2)(-1+i), \;\ (N+1/2)(1+i), \;\ (N+1/2)(-1-i), \;\ (N+1/2)(1-i)$$

The poles are located at $$z=0 (\text{order }9), \;\ z=\pm 1, \;\ \pm 2,\ldots, \;\ z=\pm i, \;\ \pm 2i,\ldots$$

So, using the series for the respective trig functions, I get:

$$\frac{\pi \cos(\pi z)\cosh(\pi z)}{z^{7}\sin(\pi z)\sinh(\pi z)}$$

$$=\pi \frac{\left(1-\frac{(\pi z)^{2}}{2!}+\frac{(\pi z)^{4}}{4!}-\cdots\right)\left(1+\frac{(\pi z)^{2}}{2!}+\frac{(\pi z)^{4}}{4!}+\cdots\right)}{z^{7}\left({\pi}z-\frac{(\pi z)^{3}}{3!}+\frac{(\pi z)^{5}}{5!}-\cdots \right)\left({\pi}z+\frac{(\pi z)^{3}}{3!}+\frac{(\pi z)^{5}}{5!}+\cdots\right)}$$

$$=\pi \frac{\left(1-\frac{(\pi z)^{4}}{6}+\cdots \right)}{z^{7}(\pi z)^{2}\left(1-\frac{(\pi z)^{4}}{90}+\cdots \right)}$$

Which leads to a residue at z=0 of $\frac{-7{\pi}^{7}}{4050}$, since this is the coefficient of the 1/z term.

The residue at $z=n$ is $\lim_{z\to n}\frac{(z-n)}{\sin(\pi z)}\cdot \frac{\pi \cos(\pi z)\coth(\pi z)}{z^{7}}=\frac{\coth(\pi n)}{n^{7}}$

The residue at $z=ni$ is $\lim_{z\to ni}\frac{(z-ni)}{\sinh(\pi z)}\cdot \frac{\pi cot(\pi z)cosh(\pi z)}{z^{7}}=\frac{coth(\pi n)}{n^{7}}$

Now, here is where I am hung up. Where does the $\frac{19}{56700}$ come from?.

There is apparently an error I am making or something I should do I am unaware of.

So, by residue theorem, I should get something like:

$$\oint_{C_{N}}\frac{\pi \cot(\pi z)\coth(\pi z)}{z^{7}}dz=\frac{-7{\pi}^{7}}{4050}+\text{something}\sum_{n=1}^{N}\frac{\coth(\pi n)}{n^{7}}$$.

What I am doing wrong or overlooking?. I do not know how to obtain the $\frac{19}{56700}$. In order to get $\frac{19}{56700}$, the $\text{something}$ would have to be $\frac{98}{19}$. I could understand it being $4\sum_{n=1}^{\infty}\frac{coth(\pi n)}{n^{7}}$. Of course, this would result in $\frac{7{\pi}^{7}}{16200}$. Help is greatly appreciated. Thanks very much.

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Interesting -- we usually have the opposite problem that people don't use double dollar signs for displayed equations and everything looks cramped :-) Note that you can get inline math by using single dollar signs; that would make things like "The residue at $z=n$ is ..." easier to read. –  joriki Apr 29 '12 at 17:17
    
Thanks for the tip, Joriki. I 'uncramped' a few things. –  Cody Apr 29 '12 at 21:34
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1 Answer

up vote 4 down vote accepted

Alternative proofs of this formula (and generalizations) may be found in :

UPDATE: concerning your method it should work since the Laurent series of your function is $$\frac 1{\pi z^9}-\frac {7\pi^3}{45z^5}-\frac{19\pi^7}{14175 z}+\operatorname{O}(z^3)$$ (the error could be in the Taylor expansion of numerator and denominator and concern the hidden coefficients...)

The $\dfrac 1z$ factor is correct up to a coefficient $4$ coming from the $\cot$ series in the four directions as required. Hoping it helped more,

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Thanks for the links. –  Cody Apr 29 '12 at 23:15
    
Thanks Raymond. I miscalculated the 1/z term when I expanded. That now makes sense. Thanks a bunch. –  Cody Apr 30 '12 at 19:15
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