Question

If $a_n=\sum \frac{n}{n!}$ then the sum of series $\sum a_n$ is

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

Answer 1

The correct option is C

$\frac{3e}{2}$

Explanation for the correct option:

Step 1. Find the value of $\sum a_n$:

Given, $a_n=\sum \frac{n}{n!}$

$$\begin{gathered} =\frac{n(n+1)}{2n!} \\ =\frac{(n+1)}{2(n-1)!} \\ =\frac{\left(n-1+2\right)}{2(n-1)!} \\ =\frac{1}{2}\left[\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\right] \end{gathered}$$

Step 2. Find the values of $a_1,a_2,a_3$ and so on:

$a_1=1$

$a_2=\left(\frac{1}{2}\right)\left(\frac{1}{0!}\right)+\frac{1}{1!}$

$a_3=\left(\frac{1}{2}\right)\left(\frac{1}{1!}\right)+\frac{1}{2!}$

$\begin{array}{rcl}\therefore \sum a_n& =& \frac{1}{2}\left[\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+....\right]+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...\\ & =& \left(\frac{1}{2}\right)e+e\\ & =& \frac{\mathbf{3}\mathbf{e}}{\mathbf{2}}\end{array}$ $\begin{array}{rcl}& & \left[\mathbf{\because }\frac{\mathbf{1}}{\mathbf{0}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{1}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{=}\mathbf{}\mathit{e}\right]\end{array}$

Hence, Option ‘C’ is Correct.