IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

the nth term is ...

$T_n=tan^{-1}\bigg(\frac{1}{1+n+n^2}\bigg)$

$T_n=tan^{-1}\bigg(\frac{n+1-n}{1+n(n+1)}\bigg)$

$T_n=tan^{-1}(n+1)-tan^{-1}n$

sum upto n terms is

$S_n=\sum_{k=1}^{n}T_k$

$S_n=\sum_{k=1}^{n}tan^{-1}(k+1)-tan^{-1}k$

$S_n=tan^{-1}(n+1)-tan^{-1}1$ (by telescopic cancellations)

$S_n=tan^{-1}\frac{n+1-1}{1+n+1}$

$S_n=tan^{-1}\frac{n}{n+2}$

sum upto infinity is

$S_{\infty}=\lim_{n\to\infty}tan^{-1}\bigg(\frac{n}{n+2}\bigg)$

$S_{\infty}=\lim_{n\to\infty}tan^{-1}\bigg(\frac{1}{1+\frac{2}{n}}\bigg)$

$S_{\infty}=tan^{-1}1$

$S_{\infty}=\frac{\pi}{4}$

hope you are clear.....