Question

Let $f\left(n\right)=\sum _{k=1}^n{\text{cosec}}^{-1}\sqrt{(k^2+1)(k^2+2k+2)}$. Then the value of $\underset{n\to \infty }{lim}\frac{12f\left(n\right)}{\pi }$ is

Answer 1

$f\left(n\right)=\sum _{k=1}^n{\text{cosec}}^{-1}\sqrt{(k^2+1)(k^2+2k+2)}$

Let $\theta ={\text{cosec}}^{-1}\sqrt{(k^2+1)(k^2+2k+2)},\theta \in (0,\frac{\pi }{2}]$
$\Rightarrow {\text{cosec}}^2\theta =(k^2+1)(k^2+2k+2)\Rightarrow {\text{cosec}}^2\theta =k^4+2k^3+2k^2+k^2+2k+2\Rightarrow {\text{cosec}}^2\theta =(k^2+k+1{)}^2+1$
$\Rightarrow {\cot }^2\theta =(k^2+k+1{)}^2$
$\Rightarrow \cot \theta =k^2+k+1(\because \theta \in (0,\frac{\pi }{2}])$
$\Rightarrow \tan \theta =\frac{1}{k^2+k+1}\Rightarrow \tan \theta =\frac{(k+1)-k}{1+(k+1)k}$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{(k+1)-k}{1+(k+1)k}\right]\Rightarrow \theta ={\tan }^{-1}(k+1)-{\tan }^{-1}\left(k\right)$

Thus, the required sum of $n$ terms of the given series is
$f\left(n\right)={\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2...+{\tan }^{-1}(n+1)-{\tan }^{-1}n$

$\Rightarrow f\left(n\right)={\tan }^{-1}(n+1)-{\tan }^{-1}1$
$\Rightarrow f\left(n\right)={\tan }^{-1}(n+1)-\frac{\pi }{4}$

$\underset{n\to \infty }{lim}\frac{12f\left(n\right)}{\pi }$
$=\frac{12}{\pi }\underset{n\to \infty }{lim}\left({\tan }^{-1}\right(n+1)-\frac{\pi }{4})$
$=\frac{12}{\pi }(\frac{\pi }{2}-\frac{\pi }{4})$
$=3$