Question

The sum of the series ${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+\cdots \text{upto}\infty$ terms is

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

Answer 1

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

We have,
${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+\cdots \text{upto}\infty$
$\therefore T_n={\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}}$
$\Rightarrow T_n={\tan }^{-1}\frac{2^{n-1}(2-1)}{1+2^n\cdot 2^{n-1}}$
Using the formula:
$\{{\tan }^{-1}\frac{x-y}{1+xy}={\tan }^{-1}x-{\tan }^{-1}y\}$
$={\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1}$
Now put $n=1,2,3,...,n$ and add, we get
$S_n={\tan }^{-1}{2}^n-{\tan }^{-1}1$
$\therefore S_{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$