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I'm trying to prove this :$$\sum_{k=1}^{m-1} \frac{1}{\sin^2\left({\frac{k\pi}{m}}\right)}=\frac{m^2-1}{3}$$ I find this Answer But I can understand this.

$$\sum_{n \in {Z}} \sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{1}{\sin^2(x)}$$

I know:$$\sum_{n \in {Z}} \frac{1}{(x+n\pi)^2}=\frac{1}{\sin^2(x)}$$

What about this is it wrong?

$$\sum_{n \in {Z}} \sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\sum_{n \in {Z}} \frac{1}{(x+(mn)\pi)^2}+\frac{1}{(x+(1+mn)\pi)^2}+...+\frac{1}{(x+(m-1+mn)\pi)}=\frac{m}{\sin^2(x)}$$

Delta
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    Hint: any integer $p$ can be written uniquely as $p=mn+k$ where $0\le k<m$. – Pranay Jan 27 '25 at 17:56
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    Yes, more generally, $$\sum_{p\in\mathbb Z} a_p = \sum_{n\in\mathbb Z}\sum_{k=0}^{m-1} a_{k+nm},$$ as long as the left side converges. You are just grouping every $m$ elements of the sum together. – Thomas Andrews Jan 27 '25 at 18:12

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More generally, $$\sum_{p\in\mathbb Z} a_p = \sum_{n\in\mathbb Z}\sum_{k=0}^{m-1} a_{k+nm},$$ as long as the left side converges.

You are just grouping every $m$ elements of the sum together.

If $m=2,$ for example. this is just: $$\cdots a_{-2}+a_{-1}+ a_0+a_1+\cdots \\=\cdots +(a_{-2}+a_{-1})+(a_0+a_1)+\cdots$$

It is possible for the right side to converge if the left side doesn't - for example, $m=2, a_p=(-1)^p.$ But you can show if $a_p\to 0$ as $p$ goes to either $\pm \infty,$ then the right side doesn't converge if the left doesn't.

Thomas Andrews
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