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The series is

$$\tan^{-1} (\frac{1}{(2)(1^2)}) + \tan^{-1} (\frac{1}{(2)(2^2)})+ \tan^{-1} (\frac{1}{(2)(3^2)}).....$$

I know the series will telescope, but I don’t know how it would work. These problems usually reduce to $\tan^{-1} a- \tan^{-1} b$, but that’s not happening here

Aditya
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    https://math.stackexchange.com/questions/1849546/evaluate-sum-n-1-infty-cot-12n2 OR https://math.stackexchange.com/questions/3053616/find-sum-n-1-infty-tan-1-frac2n2 – lab bhattacharjee Nov 20 '20 at 13:09

1 Answers1

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You can prove by induction the $n$th partial sum is $\arctan\frac{n-1}{n}$, so the series has limit $\arctan1=\frac{\pi}{4}$. So your telescoping strategy works viz. $a=\frac{k}{k+1},\,b=\frac{k-1}{k}$.

J.G.
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