Question

If $\sum _{r=1}^n{\tan }^{-1}\left(\frac{x_r-x_{r-1}}{1+x_{r-1}{x}_r}\right)=$ $\sum _{r=1}^n({\tan }^{-1}{x}_r-{\tan }^{-1}{x}_{r-1})={\tan }^{-1}{x}_n-{\tan }^{-1}{x}_0$

The sum of the series ${\tan }^{-1}\frac{1}{{2.1}^2}+{\tan }^{-1}\frac{1}{{2.2}^2}+{\tan }^{-1}\frac{1}{{2.3}^2}+......$ up to Infinite terms is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

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

${tan}^{-1}\frac{1}{2.1^2}+{tan}^{-1}\frac{1}{2.2^2}+{tan}^{-1}\frac{1}{2.3^2}+...\infty ={lim}_{n\to \infty }\sum {tan}^{-1}\frac{1}{2.r^2}={lim}_{n\to \infty }\sum {tan}^{-1}\frac{2}{4.r^2}={lim}_{n\to \infty }\sum {tan}^{-1}\frac{2}{1+4r^2-1}={lim}_{n\to \infty }\sum {tan}^{-1}\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}={lim}_{n\to \infty }({tan}^{-1}(2n+1)-{tan}^{-1}1)={lim}_{n\to \infty }\left({tan}^{-1}\left(\frac{n}{n+1}\right)\right)={lim}_{n\to \infty }\left({tan}^{-1}\left(\frac{1}{1+\frac{1}{n}}\right)\right)={tan}^{-1}1=\frac{\pi }{4}$