I marked my question in the image. My problem is that I don’t see that the two marked formula for the function $\frac{\pi}{\sin(\pi z)}$ are different.
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Where did the $(-1)^n/n$ part go? – Mark S. Apr 06 '21 at 12:18
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What textbook is this from? – Nolan P Jul 11 '21 at 19:25
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The formulas given are $$ h(z)=\frac1z+\sum_{n\ne0}\left[\frac{(-1)^n}{z-n}+\frac{(-1)^n}n\right]\tag1 $$ and $$ h(z)=\frac1z+\sum_{n=1}^\infty(-1)^n\left[\frac1{z-n}+\frac1{z+n}\right]\tag2 $$ The standard convention is that $$ \sum_{n=1}^\infty a_n=\lim_{N\to\infty}\sum_{n=1}^Na_n\tag3 $$ However, the less commonly known, but often used, convention is that for double-ended sums, $$ \sum_{n\in\mathbb{Z}}a_n=\lim\limits_{N\to\infty}\sum\limits_{n=-N}^Na_n\tag4 $$ where we maintain the standard convergence requirement that $\lim\limits_{|n|\to\infty}|a_n|=0$. Sometimes, $(4)$ is called the Principal Value of the sum, which is the same as the convention.
Applying $(4)$ to $(1)$, we get that $$ \sum_{n\ne0}\frac{(-1)^n}n=0\tag5 $$ in which case, both $(1)$ and $(2)$ agree.
robjohn
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This is perfect! My problem is that the book claims that formula (1) and formula (2) differs from each other by an intire function! I don't understand why, I think these formulas are the same, just like you wrote it. – jam Apr 06 '21 at 13:19
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$0$ is an entire function, but I myself have often used $$\pi\csc(\pi z)=\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{n+z}$$ as well as $$\pi\cot(\pi z)=\sum_{n\in\mathbb{Z}}\frac1{n+z}$$ (see this answer) – robjohn Apr 06 '21 at 13:25
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Ok, so you think that the book refers to the zero entire function in the yellowed marked sentence, don't you? – jam Apr 07 '21 at 10:09
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