What is $$\sum_{n=1}^\infty\frac{1}{n^2+1}$$ I know that $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$ but what about if we add a one to it in the denominator, can we still figure it out?
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$ \frac{1}{2} (π coth(π) - 1)$ – Tahir Imanov Oct 06 '19 at 14:33
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The correct answer is (pi - coth(pi))/2 and that question has been asked before on MSE. I just can't make the link work on my phone. If you put your question in google you can also find a generalization if you replace 1 by $a^2$ – imranfat Oct 06 '19 at 14:36
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Hint: Consider the function $f(z)=\dfrac{\pi\cot\pi z}{z^2+1}$ over $\Bbb C$ and use the Residue and Estimation Theorems from complex analysis. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Oct 06 '19 at 14:37