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.

  • $\begingroup$ 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


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


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

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


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

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

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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.