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

The correct option is B

$e^{-1}$

Explanation for the correct answer:

Let the given summation be $S=\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+.....$

The $n^{th}$ term in this summation can be written as

$t_n=\frac{2n}{\left(2n+1\right)!}$

$\Rightarrow$ $t_n=\frac{2n+1-1}{\left(2n+1\right)!}$

$\Rightarrow$ $t_n=\frac{2n+1}{\left(2n+1\right)!}-\frac{1}{\left(2n+1\right)!}$

$\Rightarrow$ $t_n=\frac{1}{\left(2n\right)!}-\frac{1}{\left(2n+1\right)!}$

$\Rightarrow$ $t_1=\frac{1}{2!}-\frac{1}{3!}$

$\Rightarrow$ $t_2=\frac{1}{4!}-\frac{1}{5!}$

$\therefore$ $S=\sum _{}{t}_n$

$\Rightarrow$ $S=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+....$

$\Rightarrow$ $S=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+....$

We know that, $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....$

Substitute $x=-1$

$e^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}+....$

$\Rightarrow$ $S=e^{-1}$

Hence, the value of $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+.....$ is $e^{-1}$.

Hence, option $\left(B\right)$ is the correct answer.