Question

The sum of the series $\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+...$ to $\infty$ is equal to

• A. $\frac{1}{2}e$
• B. $\frac{1}{2e}$
• C. $\frac{3}{2e}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{2e}$

$T_n=\frac{n}{(2n+1)!}$
$=\frac{1}{2}\left[\frac{2n+1-1}{(2n+1)!}\right]$
$=\frac{1}{2\left(2n\right)!}-\frac{1}{2(2n+1)!}$
Applying summation, we get
$=\frac{1}{2}\left[\right(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty )-(\frac{1}{3!}+\frac{1}{5!}+...\infty \left)\right]$
$=\frac{1}{2}[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+...\infty ]$
$=\frac{1}{2}.e^{-1}$
$=\frac{1}{2e}$.