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How can I evaluate the summation:

$\frac{1}{1^2+1}+\frac{1}{2^2+1}+\frac{1}{3^2+1}+\frac{1}{4^2+1}+\cdots$

I noticed that each term could be represented as $\dfrac{d(arctan(nx))}{(n)dx}=\dfrac{1}{n^2x^2+1}$

When $x=1$, the expression $\sum_{n=1}^{\infty}\dfrac{d(arctan(nx))}{ndx}$ becomes $\frac{1}{1^2+1}+\frac{1}{2^2+1}+\frac{1}{3^2+1}+\frac{1}{4^2+1}+\cdots$

However I cannot proceed any further. Is an exact value for the summation derivable?

LithiumPoisoning
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  • Can this link help you ? https://math.stackexchange.com/questions/736860/find-the-infinite-sum-of-the-series-sum-n-1-infty-frac1n2-1?noredirect=1&lq=1 – OnTheWay Aug 19 '23 at 14:26
  • Thank you, the answer is much more complicated than I can comprehend for now. Thanks anyway :) – LithiumPoisoning Aug 19 '23 at 14:30

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