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This is another 'arctanlog' series: $$ S=\sum_{n=1}^{\infty}(-1)^{n-1}\arctan\left(\frac{1}{n}\right)\ln(n^2+1) $$ Maybe differentiating with respect to some parameter could be of interest.

What method could you recommend someone to find a closed form for the above series?

Thank you.

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We have: $$ S = \frac{1}{2i}\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\log^2(n+i)-\log^2(n-i)\right)=\frac{1}{2i}\frac{d^2}{ds^2}\left.\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\frac{1}{(n+i)^s}-\frac{1}{(n-i)^s}\right)\right|_{s=0}$$ hence the original series can be written in terms of the derivatives of a Hurwitz $\zeta$-function.

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Euler's student Oct 7 '14 at 16:18

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