I asked the question at MSE 
<https://math.stackexchange.com/questions/982388/simple-finite-series-with-reciprocal-factorials>
but got no answer or comment (it is not a homework).


I'm trying to find the following sum:
$$
\sum\_{k=0}^n{1\over(n-k)!}{x^{k+2}\over k+2}.
$$
The most obvious way is to differentiate wrt to $x$ leading to 
$$
\sum\_{k=0}^n{1\over(n-k)!}{x^{k+1}}=\frac{1}{n!}e^{\frac{1}{x}} x^{n+1} \Gamma \left(n+1,\frac{1}{x}\right)
$$
(according to Mathematica) but I don't see how to integrate the right side to get the original sum ($\Gamma$ is the incomplete Gamma function).


Idea I for a solution: Isn't it possible to insert an elementary auxiliary function $f(y)$ to the first equation, whose integral wrt to $y$ will cancel the ${1\over k+2}$ term, then find the sum, differentiate it wrt to $y$ and set the auxiliary function to one? 


Such a function exists $f(y)=y^{-k/(k + 2)}$ but the sum is then unsummable.


Idea II for a solution: Multiply/divide the summands by both $n!$ and $k!$ to write it as a combinatorial series (omitting some constant functions of $n$ and $x$)
$$
\sum\_{k=0}^n\binom{n+2}{k+2}{x^{k}}k!(k+1).
$$
A lot is known about series with binomial coefficients (see here <http://www.math.wvu.edu/~gould/>) but I found no way out of it. I think the last form indicates that the sum can be written in terms of elementary functions (but it may be quite a huge expression for an arbitrary $k$).



Finite series with reciprocal factorials
According to Maple your sum is:


$$ \,{\frac {{x}^{2}{\_3F\_1(1,2,-n;\,3;\,-x)}}{2n!}} $$



```
1/2/n!*x^2*hypergeom([1, 2, -n],[3],-x)

```

