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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 !!

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

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

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

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.

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  • $\begingroup$ 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

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