Question

The sum of the series $1-\frac{1}{3.3}+\frac{1}{{5.3}^2}-\frac{1}{{7.3}^3}+...to\infty$ is

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

Answer 1

Answer: (C)

$\frac{\pi }{2\sqrt{3}}$

Gregory's series: $\Theta =\tan \Theta -\frac{1}{3}{\tan }^3\Theta +\frac{1}{5}{\tan }^5\Theta -\frac{1}{7}{\tan }^7\Theta +...\infty$ where, $\left|\Theta \right|\le \frac{\pi }{4}$
$\frac{\Theta }{\tan \Theta }=1-\frac{1}{3}{\tan }^2\Theta +\frac{1}{5}{\tan }^4\Theta -\frac{1}{7}{\tan }^6\Theta +...\infty$ ....(1)
Substitute $\tan \Theta =\frac{1}{\sqrt{3}}$ $\Rightarrow \Theta =\frac{\pi }{6}$ in (1)

$\frac{\pi \sqrt{3}}{6}=1-\frac{1}{3.3}+\frac{1}{{5.3}^2}-\frac{1}{{7.3}^3}+...\infty$
$\Rightarrow 1-\frac{1}{3.3}+\frac{1}{{5.3}^2}-\frac{1}{{7.3}^3}+...\infty =\frac{\pi }{2\sqrt{3}}$

Ans: C