Question

The sum of the series $1+\frac{1}{3}\cdot \frac{1}{4}+\frac{1}{5}\cdot \frac{1}{4^2}+\frac{1}{7}\cdot \frac{1}{4^3}+\dots \dots \infty$ is

• A. ${\log }_{e}1$
• B. ${\log }_{e}2$
• C. ${\log }_{e}3$
• D. ${\log }_{e}4$

Answer 1

Answer: (C)

Since, $\log (1+x)-\log (1-x)$
$=2\left[x+\frac{x^3}{3}+\frac{x^5}{5}+\dots \infty \right]$
Put $x=\frac{1}{2}$ on both sides, we get
$\log \left(\frac{3}{2}\right)-\log \left(\frac{1}{2}\right)$
$=2\left(\frac{1}{2}+\frac{1}{3}\cdot \frac{1}{2^3}+\frac{1}{5}\cdot \frac{1}{2^5}+...\infty \right)$
$\Rightarrow \log 3=1+\frac{1}{3}\cdot \frac{1}{4}+\frac{1}{5}\cdot \frac{1}{4^2}+\dots$