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.

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

$$5e=2+S.$$

