In reference to this question about showing that the following interesting series takes on the value

$$\sum_{n=0}^\infty \frac{1}{(2n+1)\operatorname{sinh}((2n+1)\pi)}=\frac{\log(2)}{8}$$

I tried the approach of using the residue theorem, from which one can show that

$$\sum_{n=-\infty}^\infty f(n) = -\sum_k \mathrm{Res}_{z=z_k} [\pi \cot{\pi z} \: f(z)] $$

where $z_k$ are non-integral poles of $\pi \cot{\pi z} f(z)$. In this case,

$$f(z) = \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1}$$

There is a pole of order $1$ at $z=-1/2$. Note that is is not of order $2$ because $\cot{z} \sim (\pi/2) - z$ as $z \rightarrow \pi/2$. One may then compute the single residue of $\pi \cot{\pi z} f(z)$ as

$$\mathrm{Res}_{z=z_k} \left [\pi \cot{\pi z} \: \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1} \right ] = -\frac{\pi}{4} $$

So I get as a value for the posted series as $\pi/8$, which is not correct. Nevertheless, I cannot locate my error. Perhaps someone can. For example, the series obviously converges, and quickly. But is there an error in my assumption that

$$\lim_{N \rightarrow \infty} \oint_{C_N} dz \: \pi \cot{\pi z} \: \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1} = 0$$

where $C_N$ is a rectaingle symmetric about the real and imaginary axes, extending from $[-N-1/2,N+1/2]$? Or is there something else I do not see?

  • $\begingroup$ $\sinh(z)$ has zeros on the imaginary axis. I don't see how you account for them. $\endgroup$ – Marek Feb 15 '13 at 11:08
  • $\begingroup$ I have no idea if the integral evaluates to zero on all sides of the rectangle in the limit. Well, I'm pretty sure it does on the left and right sides. But $f(z)$ has infinitely-many poles, not just one. $\endgroup$ – Random Variable Feb 15 '13 at 11:08
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    $\begingroup$ I'd like to see more such questions involving residue theorem (+1) $\endgroup$ – user 1591719 Feb 15 '13 at 11:33
  • $\begingroup$ D'oh! Of course, of course, the imaginary axis, how could I have missed that? $\endgroup$ – Ron Gordon Feb 15 '13 at 11:47
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    $\begingroup$ ...and this method simply transforms the sum into another, equally difficult sum. $\endgroup$ – Ron Gordon Feb 15 '13 at 11:57

Fulfilling joriki's request, I am providing this "no-go" answer.

The method OP wished to apply rests on the assumption (that I did not verify) that the integral around the big square $C_N$ goes to zero. If this is the case, the sum of the residues is also zero and therefore this sum can be split into two equal parts -- the one coming from $\cot(\pi z)$ representing the original sum and the part coming from the poles of the function $f(z)$ itself.

If $f(z)$ had only few singularities, we would succeed in solving the problem. This is not the case here though, since $\sinh(\pi(2z + 1)) = (-i)\sin(i\pi(2z + 1))$ has infinitely many zeros on the line $-1/2 + it$, so the resulting sum is about as complicated as the original sum. Therefore it would seem this method is not applicable for this problem.


Here is one way to consider the series.

Expand it as

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(e^{\pi n}+1)}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(e^{2\pi n}+1)}+\sum_{n=1}^{\infty}\frac{1}{n(e^{\pi n}-1)}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(e^{2\pi n}-1)}=\frac{log(2)}{8}$

These are rather famous sums studied by Ramanujan. We can probably find their evaluations in his notebooks or some other place like Plouffe, as was previously mentioned.

Can any of the above sums be evaluated using residues?.

I think there may be a general formula derived by Ramanujan for sums of this type.

I think I have seen one in terms of Bernoulli numbers.

Anyway, some thoughts.


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