Coefficient of two power series

Question

How do I find the coefficient of $x^n$ in $\sum_{k = 0}^{\infty}\frac{(-x)^k}{k!}\sum_{j = 0}^{\infty}x^j$?

Answer 1

Being lazy, I started with a small problem $$\sum_{k = 0}^{7}\frac{(-x)^k}{k!}\sum_{j = 0}^{76}x^j=1+\frac{x^2}{2}+\frac{x^3}{3}+\frac{3 x^4}{8}+\frac{11 x^5}{30}+\frac{53 x^6}{144}+\frac{103 x^7}{280}+\cdots=\sum_{k = 0}^{7}\frac{a_n}{b_n}x^n$$

Coefficients $a_n$ correspond to sequence $A053557$ in $OEIS$ (have a look here) and coefficients $b_n$ correspond to sequence $A053556$ in $OEIS$ (have a look here) making $$\frac{a_n}{b_n}=\frac{(!n)}{n!}$$ where appears the subfactorial function (also called the derangement number) (have a look here)

Answer 2

Consider the finite case.

$$\sum_0^n a_ix^i\sum_0^n b_jx^j=a_0b_0+(a_1b_0+a_0b_1)x+(a_2b_0+a_1b_1+a_2b_0)x^2+\ldots=\sum_{k=0}^{2n}\sum_{j=0}^ka_jb_{k-j}x^k$$

Can you extend it for the infinite case now?