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
    $\begingroup$ You have to write $\sum_{n=1}^N \left (\arctan (n+1)-\arctan(n-1)\right )$ $\endgroup$ – student Nov 24 '12 at 16:45

Everything is fine with this proof.

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  • $\begingroup$ Thanks, that's all I wanted to hear! $\endgroup$ – Chon Nov 24 '12 at 17:07

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