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This question already has an answer here:

The series $$\sum\limits_{n = 1}^\infty {\arctan \dfrac{2}{{{n^2}}}}$$ converges because it is asymptotic to $\dfrac{2}{n^2}$ which is convergent.

What is its sum and why?

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marked as duplicate by heropup, user7530, Guy, Brandon Carter, Avitus Apr 3 '14 at 15:44

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    $\begingroup$ The sum seems to be extremely complicated, wolframalpha.com/input/… $\endgroup$ – Guy Apr 3 '14 at 14:57
  • $\begingroup$ But wolfram's answer suggests that it can be converted into a telescopic sum, in terms of the hyperbolic functions. $\endgroup$ – Guy Apr 3 '14 at 14:58
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    $\begingroup$ @Sabyasachi: not so much. Instead, you can reduce the arctan to a difference of two arctans, but they do not telescope. I will detail below when I have some time. $\endgroup$ – Ron Gordon Apr 3 '14 at 15:05
  • $\begingroup$ @RonGordon okay thanks. I will keep a watch. $\endgroup$ – Guy Apr 3 '14 at 15:06
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    $\begingroup$ OP has changed the question. This one is a telescoping sum indeed. $\endgroup$ – Ron Gordon Apr 3 '14 at 15:22
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OK, now that you have changed the question...

$$\arctan{\frac{2}{n^2}} = \arctan{\frac1{n-1}} - \arctan{\frac1{n+1}}$$

so that the sum is

$$\frac{\pi}{2}-\arctan{\frac12} + \frac{\pi}{4}-\arctan{\frac13} + \arctan{\frac12}-\arctan{\frac14}+\cdots = \frac{3 \pi}{4}$$

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    $\begingroup$ Interesting how simply changing a $1$ to a $2$ makes everything so much simpler(even though I tend to assume that less constants usually mean less complications). Lesson in caution, if something looks simpler, it might be more complicated. +1 $\endgroup$ – Guy Apr 3 '14 at 15:36
  • $\begingroup$ This is the same result I had using Mathematica... $\endgroup$ – Raffaele Apr 4 '14 at 12:00
  • $\begingroup$ But how to you deal with the 1/0 problem? $\endgroup$ – Raffaele Apr 4 '14 at 12:34
  • $\begingroup$ @Raffaele: $$\lim_{x\to 0}\arctan{\frac1{x}} = \frac{\pi}{2}$$ $\endgroup$ – Ron Gordon Apr 4 '14 at 13:24

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