Question

Sum of series ${\cot }^{-1}\left(\frac{5}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{9}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{15}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{23}{\sqrt{3}}\right)+...$ upto infinite terms is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is B $\frac{\pi }{3}$

${\cot }^{-1}\left(\frac{5}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{9}{\sqrt{3}}\right)+{\cot }^{-1}\frac{15}{\sqrt{3}}+{\cot }^{-1}\left(\frac{23}{\sqrt{3}}\right)$+.......+ infinite terms
$={\tan }^{-1}\left(\frac{\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1+\frac{2}{\sqrt{3}}\times \frac{1}{\sqrt{3}}}\right)+{\tan }^{-1}\left(\frac{\frac{3}{\sqrt{3}}-\frac{2}{\sqrt{3}}}{1+\frac{3}{\sqrt{3}}\times \frac{2}{\sqrt{3}}}\right)+..........+{\tan }^{-1}\left(\frac{\frac{n+1}{\sqrt{3}}-\frac{n}{\sqrt{3}}}{1+\frac{n+1}{\sqrt{3}}\times \frac{n}{\sqrt{3}}}\right)+.............$
$\Rightarrow S_n={\tan }^{-1}\frac{n+1}{\sqrt{3}}-{\tan }^{-1}\frac{1}{\sqrt{3}}$
$\therefore S_{\infty }=\underset{n\to \infty }{lim}{S}_n=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$