Question

Find the sum of the infinite series
$\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+\frac{56}{5!}+....$

Answer 1

$\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+\frac{56}{5!}+...$

Now, let consider the series

$S_n=4+11+22+37+56+...+T_{n-1}+T_n$ (upto $n$ terms)

$S_{n-1}=4+11+22+37+56+...+T_{n-1}$ (upto $n-1$ terms)

Subtracting above two equation

$S_n-S_{n-1}=4+7+11+15+19+...+\left({(n-1)}^{th}term\right)$

$\Rightarrow T_n=4+[7+11+15+19+...+({(n-1)}^{th}term\left)\right]$

$=4+\left(\frac{n-1}{2}\right)[14+(n-2\left)4\right]$ $(\because T_n=S_n-S_{n-1})$

$=4+\left(\frac{n-1}{2}\right)[4n+6]$

$=4+(n-1)(2n+3)$

Now, $a_n=\frac{4+(n-1)(2n+3)}{n!}$

$=\frac{4}{n!}+\frac{2n^2+n-3}{n!}$

$=\frac{4}{n!}+\frac{2n^2-2n}{n!}+\frac{3n}{n!}-\frac{3}{n!}$

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

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

$=(e-1)+2(0+e)+3\left(e\right)$

$=6e-1$