# why $\pi \cot \pi z = \sum _{n=-\infty}^{\infty} \frac{1}{z+n}$

In prove of the above claim I need prove: $\cot (\pi\cdot z)$ is bounded for $\{y \geq 1 , 0.5 \geq x \geq -0.5\}$ In stein book a function has defined:
$$\Delta ( z ) = \pi \cot \pi z - \sum _{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$$ It has proved that $\Delta ( z )$ is entire and it's sufficient to $\Delta ( z )$ be abounded.

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The content and title are asking for separate things, which one is the question? – Arjang Jan 5 '13 at 11:52
What has to do what you ask with what you say you have to prove, anyway? Can you give some background, ideas, self effort, etc.? – DonAntonio Jan 5 '13 at 11:52
Well, you changed completely the question, yet there still remains one doubt: what has to do the proof of the equality in the title with the boundness of $\,\cot \pi z\,$ wherever? Couldn't it be that you need to prove the equality and then using it you have to prove the boundness? – DonAntonio Jan 5 '13 at 12:09
Can you exhibit a value of $z$ such that $\sum_{n=1}^\infty (1-\frac{z^2}{n^2})$ converges? – Hagen von Eitzen Jan 5 '13 at 12:26
The expression in the - now modified - title may converge as $\sum_{n=-N}^N \frac1{z+n}=\frac 1z+\sum_{n=1}^N\frac{2z}{z^2-n^2}$. But $\sum_{n=1}^\infty (1-\frac{z^2}{n^2})$ (occuring in the definition of $\Delta$ and in the original question title) will never converge as the summands tend to $1$ – Hagen von Eitzen Jan 5 '13 at 13:44