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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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,

share|cite|improve this answer
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

Your Answer


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.