Question

The sum of the series $Sum=1+{\log }_{e}2+\frac{({\log }_{e}2{)}^2}{2!}+\frac{({\log }_{e}2{)}^3}{3!}+\cdots \infty$ is

• A. $2$
• B. $1$
• C. $0$
• D. $\infty$

Answer 1

The correct option is A $2$

$Sum=1+{\log }_{e}2+\frac{({\log }_{e}2{)}^2}{2!}+\frac{({\log }_{e}2{)}^3}{3!}+\cdots \infty$
Given series is expansion of $e^x$
$=e^{{\log }_{e}2}=2$