Find the infinite sum of these factorials

The $n^\text{th}$ numerator is $n^2-2n+3$ .

We know that $\sum_{n=0}^\infty \frac{1}{n!}= e$

so let's try to manipulate the numerators to match this form.

$\begin{aligned} \sum_{n=1}^\infty \frac{n^2-2n+3}{n!} &=\sum_{n=1}^\infty \frac{n(n-1)-n+3}{n!} \\ &=\sum_{n=1}^\infty \frac{n(n-1)}{n!} - \sum_{n=1}^\infty \frac{n}{n!} +\sum_{n=1}^\infty \frac{3}{n!} \\ &=\sum_{n=2}^\infty \frac{1}{(n-2)!} - \sum_{n=1}^\infty \frac{1}{(n-1)!} +\sum_{n=1}^\infty \frac{3}{n!} \\ &=\sum_{n=0}^\infty \frac{1}{n!} -\sum_{n=0}^\infty \frac{1}{n!} +3\left[\left(\sum_{n=0}^\infty \frac{1}{n!} \right) -1\right] \\ &=\boxed{3(e-1)}\end{aligned}$

We note the $n$ th term is $a_n = \dfrac {n^2-2n+3}{n!}$ . Then we have:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {n^2-2n+3}{n!} \\ & = \sum_{n=1}^\infty \frac {n^2}{n!} - \sum_{n=1}^\infty \frac {2n}{n!} + \sum_{n=1}^\infty \frac 3{n!} \\ & = \sum_\blue{n=1}^\infty \frac n{(n-1)!} - \sum_\blue{n=1}^\infty \frac 2{(n-1)!} + 3(e-1) \\ & = \sum_\red{n=0}^\infty \frac {n+1}{n!} - \sum_\red{n=0}^\infty \frac 2{n!} + 3e-3 \\ & = \frac d{dx}\sum_{n=0}^\infty \frac {x^{n+1}}{n!} \bigg|_{x=1} - 2e + 3e-3 \\ & = \frac d{dx} xe^x \bigg|_{x=1} + e-3 \\ & = e^x + xe^x \bigg|_{x=1} + e-3 \\ & = e+e + e -3 \\ & = 3e -3 \approx \boxed{5.15} \end{aligned}$

$n th$ term or general term is $n^{2} -2n+3$

Now let's find the general upper term $2 ,3 ,6 ,11, 18,.....$ = $y$

First difference $1 ,3, 5 ,7, ......$ = $∆ y$

Second difference

$2, 2, 2,....$ = $∆² y$

So according to General formula

                                                                        ( **Newton's forward difference formula**)

$Y$ = $2+ (n-1)×∆T +((\frac{(n-1)(n-2)}{2!})×2×∆²T)$

$Y$ = $n^{2} -2 n +3$

And

$\displaystyle\sum_{n=1}^∞ \frac{n^{2}-2n+3}{n!}$

= $3(e-1)$