Question

The sum of the series ${\log }_4^{}2-{\log }_8^{}2+{\log }_{16}^{}2-...$ is equal to

• A. $e^2$
• B. ${\log }_e^{}2+1$
• C. ${\log }_e^{}3-2$
• D. $1-{\log }_e^{}2$

Answer 1

The correct option is D

$1-{\log }_e^{}2$

Explanation for the correct option:

The given series can be written as

$$\begin{gathered} {\log }_{2^2}^{}2-{\log }_{2^3}^{}2+{\log }_{2^4}^{}2... \\ =\frac{1}{2}-\frac{1}{3}+\frac{1}{4}...\left[\because {\log }_{y^n}\left(x^m\right)=\frac{m}{n}{\log }_y\left(x\right)\text{and}{\log }_{x}x=1\right] \\ =-\left[-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\right] \\ =-\left[\log (1+1)-1\right]\left[\because \log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...,\mathrm{Putting},\mathrm{x}=1\right] \\ =1-\log \left(2\right) \\ =1-{\log }_e^{}2 \end{gathered}$$

Hence, the correct answer is option (D).