Sum of series of fractions: $\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$ 
Find the sum of the series:
  $$\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$$

I am not able to understand how to proceed. The numerator terms have nothing in common which might result in an AP or GP.
Please guide me with the approach.
 A: Observe that the first order differences of the numerators, $4,6,8$ are increasing linearly (and hopefully continue doing so - there are few known terms), so that the general form of the numerators must be quadratic, $N_n:=an^2+bn+c$.
By extrapolation, the zero-th term is $2$, and as the second order difference is $2$, the general expression must be

$$N_n=n^2+bn+2.$$

From $N_1=4$ you draw

$$N_n=n^2+n+2$$ 

which fits all values.
Now the general term, a polynomial over a factorial

$$\frac{N_n}{D_n}=\frac{n^2+n+2}{n!}$$

is a little embarrassing as we don't have just inverse factorials (as in the development of $e$). Anyway, we observe the simplification
$$\frac n{n!}=\frac1{(n-1)!},$$ and (more tricky), for the quadratic term,

$$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac1{(n-2)!}+\frac1{(n-1)!}.$$

Then we use the decomposition 

$$\frac{N_n}{D_n}=\frac{n^2+n+2}{n!}=\frac1{(n-2)!}+\frac2{(n-1)!}+\frac2{n!},$$

to establish
$$\begin{matrix}
e=&&&\color{white}+\frac1{0!}&+\frac1{1!}&+\frac1{2!}&+\cdots\\
2e=&&\color{white}+\frac2{0!}&+\frac2{1!}&+\frac2{2!}&+\frac2{3!}&+\cdots\\
2e=&\frac2{0!}&+\frac2{1!}&+\frac2{2!}&+\frac2{3!}&+\frac2{4!}&+\cdots&&
\\\hline
 5e=&\frac2{0!}&+\frac4{1!}&+\frac8{2!}&+\frac{14}{3!}&+\frac{22}{4!}&+\cdots\\
\end{matrix}$$
so that in the end

$$5e=2+S.$$

A: Hint
Let $r$ denotes the index of the term, $r\geq 1$.
The numerator $a_r$ satisfies 
$$a_r=a_{r-1}+2r$$
with $a_0=2$.
Use method of difference to find the explicit expression for $a_r$.
Write you sum in summation notation and compare it to Taylor's series. You should eventually get $5e-2$ after some splitting and index changing.
A: $a_n = \frac{n^2 + n + 2}{n!}\\
\sum a_n = \sum\frac{n^2 + n + 2}{n!}\\
\sum_\limits{n=1}^\infty\frac{n^2 - n}{n!} + \sum_\limits{n=1}^\infty\frac{2n}{n!} + \sum_\limits{n=1}^\infty \frac {2}{n!}\\$
$\sum_\limits{n=1}^\infty\frac{n^2 - n}{n!} = 0 + \sum_\limits{n=2}^\infty\frac{1}{(n-2)!} = \sum_\limits{n=0}^\infty\frac{1}{n!} = e\\
\sum_\limits{n=1}^\infty\frac{2n}{n!} = 2\sum_\limits{n=1}^\infty\frac{1}{(n-1)!} = 2e\\
\sum_\limits{n=1}^\infty \frac {2}{n!} = 2e- 2$
$5e-2$
