Reverse engineering a Taylor expansion We have the sum: $$S(x) = \frac{x^4}{3(0!)} + \frac{x^5}{4(1!)} + \text{ }...$$ And we are told to sum the series to obtain a finite expression. My guess was to reverse engineer the expression in order to find a function which has $S(x)$ or some form of it as a taylor expansion. $$\frac{1}{x^4}S(x) = \frac{1}{3} +\frac{x}{4}+\frac{x^2}{5(2!)}$$ But I am not familiar with a function which increments the denominator by $1$ every time we take the derivative.

$$\begin{eqnarray*}S(x)&=&x^4\sum_{n\geq 0}\frac{x^j}{j!}\cdot\frac{1}{j+3}=x^4\sum_{n\geq 0}\frac{x^j}{j!}\int_{0}^{1}y^{j+2}\,dy=x^4\int_{0}^{1}y^2\sum_{n\geq 0}\frac{(xy)^j}{j!}\,dy\\&=&x^4\int_{0}^{1}y^2 e^{xy}\,dy=x\int_{0}^{x}z^2 e^z\,dz=x\cdot\left[\left(z^2-2z+2\right)e^{z}\right]_{0}^{x}\\&=&x\left(-2+(x^2-2x+2)e^x\right)=\color{red}{-2x+(x^3-2x^2+2x)e^x}.\end{eqnarray*}$$
