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I would like to show that:

$ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$

We have:

$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1) =-\frac{\pi}{4}+\arctan N+\arctan(N+1)\rightarrow \frac{3\pi}{4}$

Do you agree with my proof?

  • 4
    You have to write $\sum_{n=1}^N \left (\arctan (n+1)-\arctan(n-1)\right )$2012-11-24

1 Answers 1

6

Everything is fine with this proof.

  • 0
    Thanks, that's all I wanted to hear!2012-11-24