Question

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

Answer 1

$\displaystyle\sum_{n=0}^{\infty} \tan ^{-1}\left(\frac{\tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+1)(n+2)}\right)}{1+\tan ^{-1}\left(\frac{1}{n+1}\right) \cdot \tan ^{-1}\left(\frac{1}{n+2}\right)}\right)$
$=\displaystyle\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) $
$=\displaystyle\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 $
$\left[\cos ^{-1}(\cos (\pi-1))\right]=[\pi-1]=2$