Question

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

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

Answer 1

Answer: (C)

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