Question

The value of $\sum _{n=1}^{\infty }\left({\tan }^{-1}\left(\frac{n}{n+2}\right)-{\tan }^{-1}\left(\frac{n-1}{n+1}\right)\right)$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (A)

We have $T_1={\tan }^{-1}\frac{1}{3}-{\tan }^{-1}0$
$T_2={\tan }^{-1}\frac{1}{2}-{\tan }^{-1}\frac{1}{3}$
$T_3={\tan }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{1}{2}$
$⋮⋮⋮$
$T_n={\tan }^{-1}\left(\frac{n}{n+2}\right)-{\tan }^{-1}\left(\frac{n-1}{n+1}\right)$
$\therefore$ On adding all above equation, we get $S_n=\sum _{r=1}^n{T}_r={\tan }^{-1}\left(\frac{n}{n+2}\right)$
Hence $S_{\infty }=\underset{n\to \infty }{\mathrm{Limit}}{\tan }^{-1}\left(\frac{n}{n+2}\right)={\tan }^{-1}1=\frac{\pi }{4}$