Question

For any real number $x$, let $\cot ^{-1} x$ denote the unique real number $\theta$ in $(0, \pi)$ such that $\cot \theta=x$
If $\underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{k=1}^n \cot ^{-1}\left(1+k+k^2\right)=\cot ^{-1}(\alpha)+\cot ^{-1}(\beta)$, where $\alpha, \beta$ are prime numbers, then find $(\alpha+\beta)$.

Answer 1

$\displaystyle\sum_{k=1}^n \cot ^{-1}\left(1+k+k^2\right)=\tan ^{-1}(n+1)-\tan ^{-1} 1$
take $\underset{n \rightarrow \infty}{\text{Lim}} $, we get R.H.S. $=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}=\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{1}{\beta}\right)$
$\Rightarrow(\alpha-1)(\beta-1)=2 $
$\therefore \alpha=3, \beta=2 \text { (or } \alpha=2, \beta=3) $
$\therefore \text { Answer }=5 $