Question

$\sum _{n=3}^{\infty }{\tan }^{-1}\left(\frac{4}{n^2-2n-2}\right)$ is equal to

• A. $3\pi$
• B. $2\pi$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\sum _{n=3}^{\infty }{\tan }^{-1}\left(\frac{4}{n^2-2n-2}\right)$
$=\sum _{n=3}^{\infty }{\tan }^{-1}\left(\frac{4}{1+(n+1)(n-3)}\right)$
$=\sum _{n=3}^{\infty }\left({\tan }^{-1}(n+1)-{\tan }^{-1}(n-3)\right)$