Question

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

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

Answer 1

The correct option is A $\frac{\pi }{4}$

$T_n={\tan }^{-1}\frac{(n+1-1)}{(n+1)+1}-{\tan }^{-1}\frac{(n-1)}{n\cdot 1+1}{T}_n=[({\tan }^{-1}(n+1)-{\tan }^{-1}\left(1\right))-({\tan }^{-1}n-{\tan }^{-1}\left(1\right))]T_n={\tan }^{-1}(n+1)-{\tan }^{-1}n{T}_1={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)T_2={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)T_3={\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(3\right)T_n={\tan }^{-1}(n+1)-{\tan }^{-1}n{S}_n={\tan }^{-1}(n+1)-{\tan }^{-1}\left(1\right)\underset{n\to \infty }{lim}{S}_n={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$