Question

The value of $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+........$ is

• A. $e^{\frac{1}{2}}$
• B. $e^{-1}$
• C. $e$
• D. $e^{\frac{1}{3}}$

Answer 1

Answer: (B)

Given, $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\dots \infty$
The $n$th term of the above series
$T_n=\frac{2n}{(2n+1)!}=\frac{(2n+1)-1}{(2n+1)!}=\frac{2n+1}{(2n+1)!}-\frac{1}{(2n+1)!}$
$T_n=\frac{1}{(2n)!}-\frac{1}{(2n+1)!}$
$\therefore T_1=\frac{1}{2!}-\frac{1}{3!}$
$T_2=\frac{1}{4!}-\frac{1}{5!}$
$T_3=\frac{1}{6!}-\frac{1}{7!}$
$\begin{array}{ccc}.& .& .\\ .& .& .\\ .& .& .\end{array}$
Hence, $T_n=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots \infty =e^{-1}$