Question

Find the sum of the series ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{18}+{\tan }^{-1}\frac{1}{32}+\dots ..+\infty$ is

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

Answer 1

Answer: (D)

We have,
${\tan }^{-1}\left(\frac{1}{2r^2}\right)={\tan }^{-1}\left(\frac{2}{4r^2}\right)$
$={\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)={\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1)$
Thus,
$\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{2r^2}\right)=\sum _{r=1}^n\left[{\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1)\right]$
$={\tan }^{-1}(2n+1)={\tan }^{-1}(1)={\tan }^{-1}(2n+1)-\pi /4$
$\therefore \underset{n\to \infty }{\lim }\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{2r^2}\right)=\underset{n\to \infty }{\lim }\left[{\tan }^{-1}(2n+1)-\pi /4\right]$
$={\tan }^{-1}(\infty )-\pi /4=\pi /2-\pi /4=\pi /4.$