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). In the process I commuted the integral and differential with the sum operation. There are no issues with uniform convergence since the sum is finite.
An idea 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?