Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence: 
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. By multiplying both sides of our recurrence relation by $x^n$ and summing over $n$ from $0$ to $\infty$, we see that 
\begin{align}
\sum_{n=0} ^{\infty} a_{n+1} x^n = \frac{2}{3}\sum_{n=0} ^{\infty} (n+1)a_nx^n + \sum_{n=0} ^{\infty} (n+1)!x^n.
\end{align}
Using our definition of $A(x)$ we can rewrite the left hand side as
\begin{align*}
\sum_{n=0} ^{\infty} a_{n+1} x^n=\frac{A(x)-a_0}{x}.
\end{align*}
Such manipulations of the right hand side have been difficult because of the coefficients of the power series. 
Is there anyway to proceed from here, or are generating functions not suited to solve such a recurrence?

Being lazy and prove it by MI for fun.
Obviously, $a_0=5(0!)$, now assume $a_n=5n!$ for some natural number $n$, then
$$
\begin{aligned}
3 a_{n+1} &=2(n+1) 5 n !+5(n+1)_{0}^{1} \\
&=5 n !(2 n+2+n+1) \\
a_{n+1} &=5(n+1) !
\end{aligned}
$$
proved.
