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I'm trying to use the residue theorem to calculate $$\sum_{k=1}^{\infty} \dfrac{1}{(2k-1)^2}. $$ I came up with $\operatorname{Res}\left(\dfrac{\pi \cot(\pi z)}{(2z-1)^2},\frac12\right)=-\pi^2$ and $\operatorname{Res}\left(\dfrac{\pi \cot(\pi z)}{(2z-1)^2},0\right)=1$. This leaves me with $\pi^2-1=2\sum_{k=0}^{\infty}=2(\sum_{k=1}^{\infty}+1)$. Doing the algebra gives $\sum_{k=1}^{\infty}=\dfrac{\pi^2-2}{2}$, while the answer is $\dfrac{\pi^2}{8}$. I would appreciate it if anyone could help me identify where I am going wrong.

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  • $\begingroup$ See also Basel problem. $\endgroup$ – Lucian Nov 26 '15 at 4:11
  • $\begingroup$ I believe you double counted the residue at $0$ and missed a factor of $1/4$ on the residue at $1/2$. I posted an answer using both the chosen $f$ and a different choice for $f$. $\endgroup$ – Mark Viola Nov 26 '15 at 5:41
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METHOD 1:

Following the way forward in the original post, we note that

$$\text{Res}\left(\frac{\pi \cot(\pi z)}{(2z-1)^2},z=n\right)=\lim_{z\to n}\left((z-n)\frac{\pi \cot(\pi z)}{(2z-1)^2}\right)=\frac{1}{(2n-1)^2}$$

$$\text{Res}\left(\frac{\pi \cot(\pi z)}{(2z-1)^2},z=1/2\right)=\lim_{z\to 1/2}\left((z-1/2)\frac{\pi \cot(\pi z)}{(2z-1)^2}\right)=\frac{1}{(2n-1)^2}=-\frac{\pi^2}{4}$$

Then, since (as discussed in Method $2$) the sum of the residues add to zero, we have

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}}$$

where we used $\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}$.


METHOD 2:

Let $f(z)=\frac{\pi \tan(\pi z)}{z^2}$. Note that $f$ has simple poles a $z=0$ and $z=\frac{2n-1}{2}$ for all integer values of $n$. Also note that since $f$ is an odd function of $z$ and $O(z^{-2})$ for large $|z|$, we have

$$\oint_C f(z)\,dz=0$$

where $C$ is the closed contour in the upper-half plane comprised of the real axis, with infinitesimal semi-circular deformations around the poles, and the "infinite" semi-circle. Therefore, we have

$$\text{Res}\left(\frac{\pi \tan(\pi z)}{z^2},z=0\right)+\sum_{n=-\infty}^\infty \text{Res}\left(\frac{\pi \tan(\pi z)}{z^2},\frac{2n-1}{2}\right)=0 \tag 1$$

To calculate the residues at a simple poles, we evaluate the limits

$$\lim_{z\to 0}\left(z\,\frac{\pi \tan(\pi z)}{z^2}\right)=\pi^2 \tag 2$$

and

$$\lim_{z\to \frac{2n-1}{2}}\left(\left(z-\frac{2n-1}{2}\right)\,\frac{\pi \tan(\pi z)}{z^2}\right)=-\frac{1}{\left(\frac{2n-1}{2}\right)^2} \tag 3$$

Using $(2)$ and $(3)$ in $(1)$ we obtain

$$\pi^2=\sum_{n=-\infty}^\infty\frac{4}{(2n-1)^2}=8\sum_{n=1}^\infty\frac{1}{(2n-1)^2}\tag 4$$

whereupon dividing both sides of $(4)$ by $8$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}}$$

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    $\begingroup$ Nice function chosen for the contour. Nice answer. $\endgroup$ – Leucippus Nov 26 '15 at 5:35
  • $\begingroup$ @Leucippus Thank you! And great to see you're back. - Mark $\endgroup$ – Mark Viola Nov 26 '15 at 5:40
  • $\begingroup$ @user121955 Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Nov 27 '15 at 18:28

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