Question

If $\sum _{n=0}^{\infty }{\tan }^{-1}\left(\frac{{\cot }^{-1}\left(n^2+3n+3\right)}{1+{\cot }^{-1}(n+1){\cot }^{-1}(n+2)}\right)={\tan }^{-1}\left(\frac{p}{4}\right)$ then find the value of $[{\cos }^{-1}(\cos (p-$ 1)].
[Note: $[k]$ denotes greatest integer less than or equal to $k$.]

Answer 1

Answer: 2

$\sum _{n=0}^{\infty }{\tan }^{-1}\left(\frac{{\tan }^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+1)(n+2)}\right)}{1+{\tan }^{-1}\left(\frac{1}{n+1}\right)\cdot {\tan }^{-1}\left(\frac{1}{n+2}\right)}\right)$
$=\sum _{n=0}^{\infty }{\tan }^{-1}\left(\frac{{\tan }^{-1}\left(\frac{1}{n+1}\right)-{\tan }^{-1}\left(\frac{1}{n+2}\right)}{1+{\tan }^{-1}\left(\frac{1}{n+1}\right)\cdot {\tan }^{-1}\left(\frac{1}{n+2}\right)}\right)$
$=\sum _{n=0}^{\infty }\left({\tan }^{-1}\left({\tan }^{-1}\left(\frac{1}{n+1}\right)\right)-{\tan }^{-1}\left({\tan }^{-1}\left(\frac{1}{n+2}\right)\right)\right)$
$={\tan }^{-1}\left({\tan }^{-1}(1)\right)={\tan }^{-1}\left(\frac{\pi }{4}\right)$
$\therefore p=\pi$
$\left[{\cos }^{-1}(\cos (\pi -1))\right]=[\pi -1]=2$