$\displaystyle \sum_{n=1}^{\infty} an^{-1}\left(\frac{1}{n^2 + n + 1…

Question

$\displaystyle \sum_{n=1}^{\infty} 	an^{-1}\left(\frac{1}{n^2 + n + 1}ight)$ equals

Accepted Answer

D. $	frac{\pi}{4}$

Observe that $	an^{-1}	frac{1}{1 + n(n+1)} = 	an^{-1}	frac{(n+1)-n}{1+(n+1)n} = 	an^{-1}(n+1) - 	an^{-1}(n)$ (difference formula).
**So the partial sum telescopes:** $\sum_{n=1}^{N} = 	an^{-1}(N+1) - 	an^{-1}(1)$.
As $N 	o \infty$, $	an^{-1}(N+1) 	o 	frac{\pi}{2}$, giving $	frac{\pi}{2} - 	frac{\pi}{4} = 	frac{\pi}{4}$.

Answer

$	frac{\pi}{2}$

Answer

$	frac{\pi}{3}$

Answer

$	an^{-1}(2)$

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equals

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