Question

$\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+.......\mathrm{to}\infty$is

• A. $e$
• B. $2e$
• C. $3e$
• D. none of these

Answer 1

The correct option is D

none of these

Explanation for correct option:

Step-1: Finding $n^{th}$term of the given series.

Let the given series,

$\mathrm{S}=\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+.......\mathrm{to}\infty$

$n^{th}$ term for the series is $a_n=\frac{2\mathrm{n}}{\left(2\mathrm{n}+1\right)!}$

Step-2: Finding sum of infinite number of series.

$$\begin{gathered} \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}=1}^{\infty }{a}_n \\ \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}=1}^{\infty }\frac{2\mathrm{n}}{\left(2\mathrm{n}+1\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}=1}^{\infty }\frac{\left(2\mathrm{n}+1\right)-1}{\left(2\mathrm{n}+1\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}=1}^{\infty }\frac{1}{\left(2\mathrm{n}\right)!}-\sum _{\mathrm{n}=1}^{\infty }\frac{1}{\left(2\mathrm{n}+1\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\left(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.....\infty \right)-\left(\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+.....\infty \right) \\ \Rightarrow {\mathrm{S}}_{\infty }=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.......\infty \\ \Rightarrow {\mathrm{S}}_{\infty }=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.......\infty \end{gathered}$$

We know that,

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

Step-3: Put $x=1$

$\Rightarrow e^{-1}=1-\frac{1}{1!}+\frac{{\left(1\right)}^2}{2!}-\frac{{\left(1\right)}^3}{3!}+.....\infty$

Therefore,

$\Rightarrow e^{-1}=1-\frac{1}{1!}+\frac{{\left(1\right)}^2}{2!}-\frac{{\left(1\right)}^3}{3!}+.....\infty$

$\Rightarrow {\mathrm{S}}_{\infty }={\mathrm{e}}^{-1}$

Hence, correct answer is option $\left(D\right)$