# Why can we differentiate this infinite series?

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?

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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

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.