# What is the sum of $\sum\limits_{n = 1}^\infty {\arctan \dfrac{2}{{{n^2}}}}$ and why? [duplicate]

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?

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

• The sum seems to be extremely complicated, wolframalpha.com/input/… – Guy Apr 3 '14 at 14:57
• But wolfram's answer suggests that it can be converted into a telescopic sum, in terms of the hyperbolic functions. – Guy Apr 3 '14 at 14:58
• @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. – Ron Gordon Apr 3 '14 at 15:05
• @RonGordon okay thanks. I will keep a watch. – Guy Apr 3 '14 at 15:06
• OP has changed the question. This one is a telescoping sum indeed. – Ron Gordon Apr 3 '14 at 15:22

$$\arctan{\frac{2}{n^2}} = \arctan{\frac1{n-1}} - \arctan{\frac1{n+1}}$$
$$\frac{\pi}{2}-\arctan{\frac12} + \frac{\pi}{4}-\arctan{\frac13} + \arctan{\frac12}-\arctan{\frac14}+\cdots = \frac{3 \pi}{4}$$
• 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 – Guy Apr 3 '14 at 15:36
• @Raffaele: $$\lim_{x\to 0}\arctan{\frac1{x}} = \frac{\pi}{2}$$ – Ron Gordon Apr 4 '14 at 13:24