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

We have the formula: $$ \frac{1}{z}+\sum_{d=1}^{\infty}\left(\frac{1}{z-d}+\frac{1}{z+d}\right)=\pi\cot\pi z $$

Then in the book I'm reading, the author differentiate it $k-1$ times to get a formula for $$ \sum_{d\in\mathbb{Z}}\frac{1}{(z+d)^k} $$

I think I can prove that when $k\geq2$, the above series is absolutely and uniformly convergent on any compact set of $\mathbb{C}$, so we can differentiate it once and once again.

But my point is, at the very beginning, how can we differentiate the following series: $$ \frac{1}{z}+\sum_{d=1}^{\infty}\left(\frac{1}{z-d}+\frac{1}{z+d}\right) $$

It is elaborately formed to avoid convergence issue, why can we differentiate it?

share|cite|improve this question
What book is this? – Potato Jan 5 '13 at 3:28
Isn't the usual result from calculus courses that you can differentiate a converging series termwise, if the differentiated series converges uniformly? This is a corollary of the easier result that you can integrate a uniformly converging series termwise. Here you can show that the differentiated series converges uniformly on a compact set that has empty intersection with the integers. And that will be enough. – Jyrki Lahtonen Jan 5 '13 at 8:04
@Potato, GTM228, A First Course In Modular Forms – hxhxhx88 Jan 6 '13 at 1:21
@JyrkiLahtonen, I got it. thank you very much! – hxhxhx88 Jan 6 '13 at 1:26
up vote 0 down vote accepted

The series $\frac{1}{z}+(\frac{1}{z-1}+\frac{1}{z+1})+...(\frac{1}{z+n}+\frac{1}{z-n})...$ should converge for any $z\not\in \mathbb{Z}$. The individual terms' absolute value are $$|\frac{1}{z-n}+\frac{1}{z+n}|=|\frac{2z}{z^{2}-n^{2}}|=\frac{2}{|z-\frac{n^{2}}{z}|}\le \frac{2}{||z|-\frac{n^{2}}{|z|}|}$$ Assume $|z|=K$, then for large enough $n$ we have $K<\frac{n^{2}}{2K}$, so the above become $\frac{2}{\frac{n^{2}}{K}-K}\le \frac{2}{\frac{n^{2}}{2K}}=\frac{4K}{n^{2}}$. So the series is absolutely convergent for any $z$, and one should be able to prove locally $f(z)$ is holomorphic by dominated convergence theorem.

share|cite|improve this answer
Oh it is. thank you! – hxhxhx88 Jan 6 '13 at 1:26

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.