Question

The sum of the infinite series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\frac{8}{9!}+\cdots \infty$ is

• A. $e$
• B. $e^{-1}$
• C. $2e$
• D. $e^2$

Answer 1

The correct option is B $e^{-1}$

$\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\frac{8}{9!}+\cdots \infty$

Let $T_n$ be the $n$th term of the given series. Then,
$T_n=\frac{2n}{(2n+1)!},n=1,2,3,\dots \infty$
$\therefore (\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots \infty )$
$=\sum _{n=1}^{\infty }{T}_n=\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}=\sum _{n=1}^{\infty }\frac{(2n+1-1)}{(2n+1)!}$
$=\sum _{n=1}^{\infty }\frac{(2n+1)}{(2n+1)!}-\sum _{n=1}^{\infty }\frac{1}{(2n+1)!}$
$=\sum _{n=1}^{\infty }\frac{1}{\left(2n\right)!}-\sum _{n=1}^{\infty }\frac{1}{(2n+1)!}$
$=(\frac{e+e^{-1}}{2}-1)-(\frac{e-e^{-1}}{2}-1)=e^{-1}$