I'm having trouble finding the sequence generated by this function.

Question

$f(x)=\frac{1}{e^x(1-x)}$

I'm aware that $e^x$ generates $1,\frac{1}{1!},\frac{2}{2!},\frac{3}{3!}...$

And I think that $\frac{e^x}{(1-x)}$ generates $a_n=\sum_{i=0}^{n}{\frac{1}{i!}}$

Answer 1

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\\\frac1{1-x}=\sum_{n=0}^\infty x^n,|x|<1$$

Thus, the Maclaurin series of $f(x)$ is $$f(x)=\frac{e^{-x}}{1-x}=\left(\sum_{n=0}^\infty\frac{(-x)^n}{n!}\right)\left(\sum_{n=0}^\infty x^n\right)\\=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{(-1)^k}{k!}\right)x^n,|x|<1$$and the required sequence is $\left\{\left(\sum_{k=0}^n\frac{(-1)^k}{k!}\right)\right\}_{n=0}^\infty$.