Let $a_0 = 0$, and let $a_{n+1} = (n-1)a_n + n!$ for $n \ge 0$. Find an explicit formula for $a_n$.
I have gotten to the point where I have $\sum_{n \ge 0}a_{n+1}\frac{x^{n+1}}{(n+1)!}=\sum_{n \ge 0}a_{n}\frac{x^{n+1}}{n!}+\sum_{n \ge 0}\frac{x^{n+1}}{n+1}$.
I also set up the exponential generating function as $A(x) = \sum_{n \ge 0}a_n\frac{x^n}{n!}$, which is used to rewrite the above as $A(x)[1-x]=-ln(1-x)$ or $A(x)=-\frac{ln(1-x)}{1-x}$.
This is where I am stuck trying to find the explicit formula for $a_n$. Can you help me find the power series representation of the right-hand side? If so, I should be able to finish the question.
I’m assuming that the recurrence is actually $a_{n+1}=(n+1)a_n+n!$, in order to justify your equation
$$\sum_{n\ge 0}a_{n+1}\frac{x^{n+1}}{(n+1)!}=\sum_{n\ge 0}a_n\frac{x^{n+1}}{n!}+\sum_{n\ge 0}\frac{x^{n+1}}{n+1}\;.$$
Taking $A(x)$ to be the generating function, I also get
$$A(x)=-\frac{\ln(1-x)}{1-x}=\frac1{1-x}\cdot\ln\frac1{1-x}=\sum_{n\ge 0}x^n\sum_{n\ge 1}\frac{x^n}n\;.$$
Now take the product:
$$\sum_{n\ge 0}x^n\sum_{n\ge 1}\frac{x^n}n=\sum_{n\ge 1}\sum_{k=1}^n\frac1kx^n=\sum_{n\ge 1}H_nx^n\;,$$
where $H_n=\sum_{k=1}^n\frac1k$ is the $n$-th harmonic number. Finally,
$$\sum_{n\ge 1}H_nx^n=\sum_{n\ge 1}n!H_n\frac{x^n}{n!}\;,$$
so $$a_n=n!H_n=n!\sum_{k=1}^n\frac1k\;.$$
In fact $\displaystyle a_n={{n+1}\brack 2}$, a Stirling number of the first kind, and the recurrence is a special case of the general recurrence
$${{n+1}\brack k}=n{n\brack k}+{n\brack{k-1}}\;,$$
since $\displaystyle{n\brack 1}=(n-1)!$.