Question

The sum of the infinite series $co{t}^{-1}2+co{t}^{-1}8+co{t}^{-1}18+co{t}^{-1}32+....$ is

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

Answer 1

Answer: (C)

Let $S_{\infty }=co{t}^{-1}2+co{t}^{-1}8+co{t}^{-1}18+co{t}^{-1}32+...$
$\therefore T_n=co{t}^{-1}2n^2$
$=ta{n}^{-1}\frac{1}{2n^2}$
$=ta{n}^{-1}\left(\frac{2}{4n^2}\right)=ta{n}^{-1}\left(\frac{\left(2n+1\right)-\left(2n-1\right)}{1+\left(2n+1\right)\left(2n-1\right)}\right)$
$=ta{n}^{-1}\left(2n+1\right)-ta{n}^{-1}\left(2n-1\right)$
$\therefore S_n=$ $\sum _{n=1}^{\infty }$$\left\{ta{n}^{-1}\left(2n+1\right)-ta{n}^{-1}\left(2n-1\right)\right\}$
$=ta{n}^{-1}\infty -ta{n}^{-1}1$
$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$