Question

The value of $\frac{1}{2!}+\frac{1+2}{3!}+\frac{1+2+3}{4!}+\dots \dots \dots ..\infty$ is

• A. $\frac{e}{2}$
• B. $\frac{e}{3}$
• C. $\frac{e}{4}$
• D. $\frac{e}{5}$

Answer 1

The correct option is A $\frac{e}{2}$

From the given series we can conclude that

$n^{th}$ term$=\frac{1+2+3+---n-1}{n!}$

${\sum }_{n=2}^{\infty }\frac{1+2+3+---n-1}{n!}$

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

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

$=\frac{1}{2}e$