Question

The sum of the series $1+{\log }_{e}2+\frac{{\left({\log }_{e}2\right)}^2}{2!}+\frac{{\left({\log }_{e}2\right)}^3}{3!}+...$to $\infty$ is

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

Answer 1

The correct option is A $2$

$1+{\log }_{e}2+\frac{{\left({\log }_{e}2\right)}^2}{2!}+\frac{{\left({\log }_{e}2\right)}^3}{3!}+....\infty$

Let $x={\log }_{e}2$

Now, the series is

$1+x+\frac{{\left(x\right)}^2}{2!}+\frac{{\left(x\right)}^3}{3!}+....\infty$

$=e^x$

$=e^{{\log }_{e}2}$

$=2$