Question

The sum of series $1+\left(\frac{1}{3}\right).\left(\frac{1}{4}\right)+\left(\frac{1}{5}\right).\left(\frac{1}{4^2}\right)+\left(\frac{1}{7}\right).\left(\frac{1}{4^3}\right)+\dots .$is

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

Answer 1

The correct option is C

${\log }_{e}3$

Explanation for the correct option:

Find the required sum.

Given: series $1+\left(\frac{1}{3}\right).\left(\frac{1}{4}\right)+\left(\frac{1}{5}\right).\left(\frac{1}{4^2}\right)+\left(\frac{1}{7}\right).\left(\frac{1}{4^3}\right)+\dots .$

We know that, $\log (1+x)=x–\frac{x^2}{2}+\frac{x^3}{3}–\frac{x^4}{4}+\dots \dots$

and $\log (1-x)=-x–\frac{x^2}{2}–\frac{x^3}{3}–\frac{x^4}{4}–\dots \dots$

So,

$$\begin{gathered} \log (1+x)-\log (1-x)=\left[x–\frac{x^2}{2}+\frac{x^3}{3}–\frac{x^4}{4}+\dots \dots \right]-\left[-x–\frac{x^2}{2}–\frac{x^3}{3}–\frac{x^4}{4}–\dots \dots \right] \\ \Rightarrow \log (1+x)-\log (1-x)=\left[x–\frac{x^2}{2}+\frac{x^3}{3}–\frac{x^4}{4}+\dots \dots +x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots \dots \right] \\ \Rightarrow \log (1+x)-\log (1-x)=\left[2x+2\frac{x^3}{3}+2\frac{x^5}{5}\dots \dots \right] \\ \Rightarrow \log (1+x)-\log (1-x)=2\left[x+\frac{x^3}{3}+\frac{x^5}{5}\dots \dots \right] \end{gathered}$$

Put $x=\frac{1}{2}$ on both sides, we get

$$\begin{gathered} \Rightarrow \log \left(1+\frac{1}{2}\right)-\log \left(1-\frac{1}{2}\right)=2\left[\frac{1}{2}+\frac{{\left(\frac{1}{2}\right)}^3}{3}+\frac{{\left(\frac{1}{2}\right)}^5}{5}+\dots \dots \right] \\ \Rightarrow \log \left(\frac{3}{2}\right)-\log \left(\frac{1}{2}\right)=2\left[\frac{1}{2}+\left(\frac{1}{3}\right)\left(\frac{1}{2^3}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{2^5}\right)+\dots \dots \right] \\ \Rightarrow \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)=\left[1+\left(\frac{1}{3}\right)\left(\frac{1}{2^2}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{2^4}\right)+\dots \dots \right]\left[\because \log m-\log n=\frac{\log m}{\log n}\right] \\ \Rightarrow {\log }_{e}3=\left[1+\left(\frac{1}{3}\right)\left(\frac{1}{4}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{4^2}\right)+\dots \dots \right] \end{gathered}$$

Hence, option (C) is the correct answer.