A closed form for $\sum_{n=1}^{\infty}(-1)^{n-1}\arctan\left(\frac{1}{n}\right)\ln(n^2+1) $

Question

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.

Answer 1

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.