Question

$\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+....$is equal to

• A. $e$
• B. $\frac{e^{-1}}{2}$
• C. $\frac{e}{4}$
• D. $\frac{e}{6}$

Answer 1

The correct option is B

$\frac{e^{-1}}{2}$

Explanation for the correct option:

Step 1. Find the sum of the given series:

Given $\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+\dots .$

$=\frac{0}{1!}+\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+\dots .$

$$$S={\displaystyle \sum _{n=0}^{\infty }}\frac{n}{(2n+1)!}$

Multiply and divide both numerator and denominator by $2$, also add and subtract $1$ from numerator

$S=\frac{1}{2}\sum _{n=0}^{\infty }\frac{(2n+1-1)}{(2n+1)!}$

$=\frac{1}{2}\left[\sum _{n=0}^{\infty }\frac{(2n+1)}{(2n+1)!}–\sum _{n=0}^{\infty }\frac{1}{(2n+1)!}\right]$

$=\frac{1}{2}\left[\sum _{n=0}^{\infty }\frac{1}{2n!}–\sum _{n=o}^{\infty }\frac{1}{(2n+1)!}\right]$

We know, $\mathit{e}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}\mathbf{!}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{\infty }$ ..….(i)

${\mathit{e}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{1}\mathbf{!}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{!}}\mathbf{–}\frac{\mathbf{1}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{\infty }$ ……(ii)

Step 2. Add equation (i) and (ii) and Subtract equation (iii) from equation (i):

$(e+e^{-1})=2\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}$ ……(iii)

and $(e-e^{-1})=2\sum _{n=0}^{\infty }\frac{1}{(2n+1)!}$ ……(iv)

Step 3. Substitute equation (iii) and (iv) in $S$:

$S=\frac{1}{2}\left[\sum _{n=0}^{\infty }\frac{1}{2n!}–\sum _{n=0}^{\infty }\frac{1}{(2n+1)!}\right]$

$=\frac{1}{2}\left[\frac{(e+e^{-1})}{2}–\frac{(e-e^{-1})}{2}\right]$

$=\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{1}}}{\mathbf{2}}$

Hence, option ‘B’ is Correct.