The value of $\sum_{r = 0}^{\infty}{ an^{- 1}\left( \frac{1}{1 + r + r^{2}} ight)}\$is equal to

Question

The value of $\sum_{r = 0}^{\infty}{	an^{- 1}\left( \frac{1}{1 + r + r^{2}} ight)}\$is equal to

Dr. Abubacker · Accepted Answer

(a)
$\because	an^{- 1}\left( \frac{1}{1 + r + r^{2}} ight) = 	an^{- 1}\left( \frac{r + 1 - r}{1 + r(r + 1)} ight)$
$= 	an^{- 1}{(r + 1) - 	an^{- 1}{(r)}}$
$	herefore\sum_{r = 0}^{n}{\lbrack	an^{- 1}{(r + 1) - 	an^{- 1}{(r)brack}}}$
$= 	an^{- 1}{(n + 1) - 	an^{- 1}(0)}$
$= 	an^{- 1}{(n + 1)}$
$\Rightarrow \sum_{r = 0}^{\infty}{	an^{- 1}{\left( \frac{1}{1 + r + r^{2}} ight) = 	an^{- 1}{(\infty) = \frac{\pi}{2}}}}$

Q. The value of $\sum_{r = 0}^{\infty}{\tan^{- 1}\left( \frac{1}{1 + r + r^{2}} \right)}\$is equal to

Explanation

Reviewed by Dr. Abubacker

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