Question

${\sum }_{n=1}^{\infty }\frac{\tan \left(\frac{2n-1}{4}\right)\pi }{2^{n+1}}=$

• A. $\frac{1}{6}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A

$\frac{1}{6}$

${\sum }_{n=1}^{\infty }\frac{\tan \left(\frac{2n-1}{4}\right)\pi }{2^{n+1}}=\frac{\tan \frac{\pi }{4}}{4}+\frac{\tan \frac{3\pi }{4}}{8}+\frac{\tan \frac{5\pi }{4}}{16}+\frac{\tan \frac{7\pi }{4}}{32}+\dots$

$=\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\dots$

This is a Geometric Progression with first term as $\frac{1}{4}$ and common ratio as $-\frac{1}{2}$

$S_{\infty }=\frac{\frac{1}{4}}{1+\frac{1}{2}}=\frac{1}{6}$