a cute problem on infinite series

Evaluate the infinite series $S = \sum_{n=1}^{\infty} \arctan (2/n^{2})$

Solution:

For $n \geq 1$ we have

$\arctan (1/n)-\arctan (1/(n+2))=\arctan \frac{(1/n)-(1/(n+2)}{1+(1/(n(n+2)))}=\arctan (2/(n+1)^{2})$

so that for $N \geq 2$ we have

$\sum_{n=2}^{N} \arctan (2/n^{2}) = \sum_{n=1}^{N-1}\arctan(2/(n+1)^{2})$ which equals the following

$\sum_{n=1}^{N-1} (\arctan (1/n) - \arctan(1/(n+2)) = \arctan (1) + \arctan (1/2) - \arctan (1/N) - \arctan (1/(N+1))$

Letting $N \rightarrow \infty$ we get

$\sum_{n=2}^{\infty} \arctan (2/n^{2}) = \arctan (1) + \arctan (1/2) = (\pi/4) + \arctan (1/2)$ and so

$S= (\pi/4) + \arctan (2) + \arctan (1/2) = ((3\pi)/4)$

More later…

Nalin Pithwa

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