TITLE: sum of series involving coth using complex analysis
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: Alternative proofs of this formula (and generalizations) may be found in :

Sandham's paper of 1954 'Some infinite series' (see (1.42) and (6.32)) 
Berndt's paper of 1977 'Modular transformations and generalizations of several formulae of Ramanujan' (page 155)
(it appears too shortly in his Ramanujan's Notebooks II (25.2))

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,