Find the value of $A=\sum_{n=-\infty}^{\infty}\frac{1}{(n-m)^2}$

Question

Find the value of $A$ where $$A=\sum_{n=-\infty}^{\infty}\frac{1}{(n-m)^2}$$ and $m\in\mathbb{Q}\setminus\mathbb{Z}$

I tried putting $t=n-m$ and rewrote the summation as $$\sum_{t=-\infty-\{0\}}^{\infty}\frac{1}{t^2}$$ But this doesn't feel right to me. Also as $m$ isn't an integer, we won't be able to use any concept of Basel problem.

Any help is greatly appreciated.

Note that $t$ does not run through the integers. It runs through some non-integral rationals. — Tzimmo, Mar 31 '25 at 13:50
The notation for your sum just looks very wrong and doesn't really capture, what values $t$ takes. Maybe sum over $t \in \mathbb{Z} - m$. — Tzimmo, Mar 31 '25 at 13:55
Do you have the answer? Is it $\pi ^2 \operatorname{cosec} ^2 (\pi m)$? — Random Math Enthusiast, Mar 31 '25 at 14:01
Are you looking for $\displaystyle\frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$? — , Mar 31 '25 at 14:02
Why do you put all those extra conditions on $m?$" Just write $m\in\mathbb Q\setminus \mathbb Z.$ — Thomas Andrews, Mar 31 '25 at 14:42
$\sum_{t=-\infty}^{\infty}$ means $t$ runs overall integers, but $t=n-m$ is never an integer, since $n$ is an integer and $m$ is not. — Thomas Andrews, Mar 31 '25 at 14:44
If $m\not\in \Bbb Z$ why you later specify $n\ne m$ and $m\ne 0$? — jjagmath, Mar 31 '25 at 14:49

score 3 · Accepted Answer · answered Mar 31 '25 at 14:17

3

we have : $$\frac{1}{\sin^2(x)}=\sum_{n=-\infty}^{\infty}\frac{1}{(x-n\pi)^2}$$

Let :$$x=m\pi$$

answered Mar 31 '25 at 14:17

Delta

1

Find the value of $A=\sum_{n=-\infty}^{\infty}\frac{1}{(n-m)^2}$

1 Answers1