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I know that $$ \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}\quad y\quad \sum_{n=0}^{\infty}\dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{8} $$ I would like to know if there is a closed formula for $f(\alpha)=\displaystyle\sum_{n=0}^{\infty}\dfrac{1}{(\alpha\cdot n+1)^2}$, since $f(1)=\dfrac{\pi^2}{6}$ and $f(2)=\dfrac{\pi^2}{8}$. If so, how could I prove it, I have been trying with $\displaystyle\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)^2}$ but without success. As mentioned in a comment, the digamma function is related but I would like to see details or references of said property, if possible.

Zaragosa
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    If $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ (that is, $\psi$ is the digamma function), then$$\sum_{n=0}^\infty\frac1{(\alpha n+1)^2}=\frac1{\alpha^2}\psi'\left(\frac1\alpha\right).$$ – José Carlos Santos Jan 01 '24 at 18:43
  • @JoséCarlosSantos Interesting, I will try to test that property you mention but it would be very helpful if you could comment or give some idea. Thank you. – Zaragosa Jan 01 '24 at 19:05

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