Question

The sum of the series $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\frac{1+2+3+4}{4!}+...\infty$ is $\frac{ae}{b}$ Where $a$ and $b$ do not have a common factor other than $1$. Find $a+b$

Answer 1

$T_n=\frac{\frac{n}{2}(2\times 1+(n-1)\times 1)}{n!}=\frac{1}{2}\frac{n(n+1)}{n!}=\frac{1}{2}\frac{n+1}{(n-1)!}$

$=\frac{1}{2}\frac{(n-1)+2}{(n-1)!}=\frac{1}{2}\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$

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

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

Therefore $a+b=5$