Wow series

Nice question.. :-)
$\Large\mathfrak{T}=\displaystyle\sum_{n=1}^{\infty}\dfrac{3n^2+n-2}{n!}$ $\large=\displaystyle\sum_{n=1}^{\infty}(\dfrac{3}{(n-2)!})+(\dfrac{4}{(n-1)!})-(\dfrac{2}{n!})$ $\Large =3(e)+4(e)-2(e-1)~~~~(\color{#302B94}{**})$ $\Large =2+5e\approx\boxed{\color{#007fff}{15.591}}$

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$\boxed{\color{#D61F06}{\large\mathcal{NOTE:}}\\\boxed{(1). \small{\text{Finding General term for numerator}(a_n)\\ \mathfrak{S}=2+12+28+\cdots+a_n~~~~~\\ \mathfrak{S}=~~~~~~~~2+12+28+\cdots+a_n \\ \text{Subtracting we get :}\\ 0=-2+\color{#20A900}{(4+10+16\cdots\text{n terms}})- a_n\\\Rightarrow a_n=-2+\color{#20A900}{3n^2+n}=3n^2+n-2}}\\\boxed{(2).(\color{#302B94}{**}) \text{Use expansion of } e^x \text{ at x=1}~:\\ e=1+\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n!} }}$

$\begin{array} {cccc} n & a_n & \Delta & \Delta^2 \\ \hline 1 & 2 \\ & & 10 \\ 2 & 12 & & 6 \\ & & 16 \\ 3 & 28 & & 6 \\ & & 22 \\ 4 & 50 \end{array}$

We note that the second difference $\Delta^2=6$ of the sequence is a constant. Then the formula of $\{a_n\}$ is of the form $a_n = an^2 + bn+c$ .

Solving $\begin{cases} n=1: & a+b+c = 2 \\ n=2: & 4a+2b+c = 12 \\ n=3: & 9a+3b+c = 28 \end{cases}$ $\implies a_n = 3n^2 + n - 2$ .

Therefore the sum is

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {3n^2+n-2}{n!} \\ & = \sum_{n=1}^\infty \frac {3n^2+n}{n!} - \sum_{n=1}^\infty \frac 2{n!} \\ & = \sum_{n=\red 0}^\infty \frac {3(n+1)^2+n+1}{(n+1)!} - \sum_{n=\blue 0}^\infty \frac 2{n!} + \blue 2 \\ & = \sum_{n=\red 0}^\infty \frac {3(n+1)+1}{n!} - 2e + 2 \\ & = \sum_{n=\red 0}^\infty \frac {3n}{n!} + \sum_{n=\red 0}^\infty \frac 4{n!} - 2e + \\ & = 3\sum_{n=\blue 1}^\infty \frac n{n!} + 4e - 2e + 2 \\ & = 3\sum_{n=\blue 1}^\infty \frac 1{(n-1)!} + 4e - 2e + 2 \\ & = 3\sum_{n=\red 0}^\infty \frac 1{n!} + 4e - 2e + 2 \\ & = 3e + 4e - 2e + 2 = 5e + 2 \approx \boxed{15.591} \end{aligned}$