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.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.