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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1The sum seems to be extremely complicated, http://www.wolframalpha.com/input/?i=sum%28arctan%281%2Fn%5E2%29%29+from+n%3D1+to+infinity – Guy Apr 03 '14 at 14:57
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But wolfram's answer suggests that it can be converted into a telescopic sum, in terms of the hyperbolic functions. – Guy Apr 03 '14 at 14:58
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2@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 03 '14 at 15:05
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@RonGordon okay thanks. I will keep a watch. – Guy Apr 03 '14 at 15:06
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'tis a duplicate indeed, although I have a different approach than those shown. I will document my approach either here or in the other venue. – Ron Gordon Apr 03 '14 at 15:10
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Ah right. Voted to close as duplicate. This might get closed, so I will keep a watch on both. – Guy Apr 03 '14 at 15:20
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1OP has changed the question. This one is a telescoping sum indeed. – Ron Gordon Apr 03 '14 at 15:22
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@RonGordon yeah. Will you still post your alternate solution on the other thread? – Guy Apr 03 '14 at 15:36
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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
$$\frac{\pi}{2}-\arctan{\frac12} + \frac{\pi}{4}-\arctan{\frac13} + \arctan{\frac12}-\arctan{\frac14}+\cdots = \frac{3 \pi}{4}$$
Ron Gordon
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1Interesting 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 03 '14 at 15:36
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