Question

${\tan }^{-1}\frac{1}{3}+ta{n}^{-1}\frac{2}{9}+ta{n}^{-1}\frac{4}{33}+....\infty$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. none

Answer 1

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

The given series can be rewritten as
$ta{n}^{-1}\frac{2^0}{1+2^1}+ta{n}^{-1}\frac{2^1}{1+2^3}+ta{n}^{-1}\frac{2^2}{1+2^5}+....$
$\therefore T_n=ta{n}^{-1}\frac{2^{n-1}}{1+2^{2n-1}}=ta{n}^{-1}\frac{2^n-2^{n-1}}{1+2^n.2^{2n-1}}$
or $T_n=ta{n}^{-1}{2}^n-ta{n}^{-1}{2}^{n-1}$
Now put n = 1, 2, 3,..., and add. The terms cancel diagonally
$\therefore S_n=ta{n}^{-1}{2}^n-ta{n}^{-1}1$
$\therefore S_{\infty }=ta{n}^{-1}\left(\infty \right)-ta{n}^{-1}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\to \left(a\right)$