Solving an exercise today I came across this series and I'm curious to know if we can evaluate it. Here it is:

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}}$$

It rings me bells about some other series with arctan's I have come across but I could not see any similarity on how to begin. Wolfram gives an approximation of $1.41379$. Note that $\sqrt{2} \approx 1.4142$. Too sad !!

  • $\begingroup$ @MartinSleziak thanks for adding the series at the title of the post $\endgroup$ – Tolaso Jan 18 '17 at 17:49


$$F(a) = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{a}{n \sqrt{n}}$$

$$F'(a) = \sum_{n=1}^{\infty} \frac{n}{a^2+n^3}$$

Using wolfram we get

$$F'(a) = -\frac{1}{3} \sum \frac{\psi(-\omega)}{\omega+1}$$

Which is summed over the roots of the equation $$\omega^3+3\omega^2+3\omega+1+a^2=0$$

I am not sure of the complexity of finding the anti derivative.

  • $\begingroup$ And if you do find the derivative integrating back seems a pain for me. Anyway, thanks Zaid! $\endgroup$ – Tolaso Jan 18 '17 at 14:36
  • $\begingroup$ @Tolaso, it is interesting that the derivative has a closed form in terms of the polygamma. $\endgroup$ – Zaid Alyafeai Jan 18 '17 at 14:40
  • $\begingroup$ Did you find an explicit closed form for the derivative ? $\endgroup$ – Tolaso Jan 18 '17 at 17:48
  • $\begingroup$ @Tolaso, The derivative is in terms of the digamma function as I illustrated. $\endgroup$ – Zaid Alyafeai Jan 19 '17 at 1:24
  • $\begingroup$ Yes ... I had seen that.. thought you made progress and simplified it in a much better form ... $\endgroup$ – Tolaso Jan 19 '17 at 7:40

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.