Question

The sum of series ${\sum }_{n=1}^{\infty }\frac{2n}{(2n+1)!}$ is

• A. $e$
• B. $e^{-1}$
• C. $2e$
• D. None of these

Answer 1

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

We have $\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}$

$=\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}=\sum _{n=1}^{\infty }\frac{2n+1-1}{(2n+1)!}=\sum _{n=1}^{\infty }(\frac{1}{\left(2n\right)!}-\frac{1}{(2n+1)!})$

$=\sum _{n=1}^{\infty }\frac{1}{\left(2n\right)!}-\sum _{n=1}^{\infty }\frac{1}{(2n+1)!}=[\frac{e+e^{-1}}{2}-1]-[\frac{e-e^{-1}}{2}-1]=e^{-1}$