Question

$\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+..........\infty$equals

• A. $e$
• B. $-e$
• C. $\frac{1}{e}$
• D. none of these

Answer 1

The correct option is A

$e$

Explanation for correct answer:

Let the given sequence, $\mathrm{S}=\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+..........\infty$

$$\begin{gathered} \Rightarrow {\mathrm{S}}_{\infty }=\sum _{n\mathit{=}\mathit{1}}^{\infty }\frac{2\mathrm{n}}{\left(2\mathrm{n}-1\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}\mathit{=}\mathit{1}}^{\infty }\frac{2\mathrm{n}-1+1}{\left(2\mathrm{n}-1\right)\left(2\mathrm{n}-2\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}\mathit{=}\mathit{1}}^{\infty }\frac{1}{\left(2\mathrm{n}-2\right)!}+\frac{1}{\left(2\mathrm{n}-1\right)!} \\ \Rightarrow {\mathrm{S}}_{\infty }=\left(\frac{1}{0!}+\frac{1}{2!}+\frac{1}{4!}+....\right)+\left(\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+......\right) \\ \Rightarrow {\mathrm{S}}_{\infty }=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.......\right) \\ \Rightarrow {\mathrm{S}}_{\infty }=\mathrm{e}\left[\because e=1+\frac{1}{1!}+\frac{1}{2!}+........\right] \end{gathered}$$

Therefore, correct answer is option $\left(a\right)$.