$\pi \cot\pi z\ne -\frac{1}{z}+\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{z+n}$

Question

I know that $$\pi\cot\pi z=\frac{1}{z}+\displaystyle\underbrace{\sum_{n=1}^\infty \left(\frac{1}{z-n}+\frac{1}{z+n}\right)}_{S}.$$ Then $$\begin{align*}S&=\cdots +\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z-2}+\frac{1}{z+2}+\cdots\\&=\cdots +\frac{1}{z-2}+\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots\\&=\cdots +\frac{1}{z-2}+\frac{1}{z-1}+\frac{1}{z-0}+\frac{1}{z+0}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots -\frac{2}{z}\\&=\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}-\frac{2}{z}.\end{align*}$$ Therefore, $$\begin{align*}\pi\cot\pi z&=\frac{1}{z}+\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}-\frac{2}{z}\\&=-\frac{1}{z}+\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}.\end{align*}$$ But the correct result is $$\pi\cot\pi z=\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}.$$ Where is the mistake?

Answer 1

Note that$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$instead of$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$which is what you wrote.