Can this recurrence relation be solved with generating functions? I have this recurrence relation,
$$a_{n+1}=\frac{n+2}{n}a_n$$
with $a_1=1$.
I've already solved this using a substitution approach by letting $a_n=\dfrac{(n+1)!}{(n-1)!}b_n$. This means $a_{n+1}=\dfrac{(n+2)!}{n!}b_{n+1}$, and so
$$\begin{align*}
\frac{(n+2)!}{n!}b_{n+1}&=\frac{n+2}{n}\frac{(n+1)!}{(n-1)!}b_n\\
b_{n+1}&=b_n\\
\sum_{n=1}^{k-1}(b_{n+1}-b_n)&=0\\
b_k&=b_1\\
b_k&=\frac{(1-1)!}{(1+1)!}a_1\\
b_k&=\frac{1}{2}
\end{align*}$$
Finally, $a_n=\dfrac{n(n+1)}{2}$.
I'm wondering if there's a way to use generating functions to solve this relation? I've been having trouble working with the $\dfrac{2}{n}a_n$ term. Are there only certain cases of relations for which generating functions are useful?

Induction is easier, but to use generating functions let's assume that
$$
f(x)=\sum_{n=1}^\infty a_nx^n
$$
Then
$$
\begin{align}
\sum_{n=1}^\infty na_{n+1}x^n
&=x\left(\frac{f(x)}x\right)'\\
&=f'(x)-\frac{f(x)}x
\end{align}
$$
and
$$
\begin{align}
\sum_{n=1}^\infty(n+2)a_nx^n
&=\frac1x\left(f(x)x^2\right)'\\
&=2f(x)+xf'(x)
\end{align}
$$
Equating these functions gives
$$
(1-x)f'(x)=\left(2+\frac1x\right)f(x)
$$
Therefore, dividing both sides by $(1-x)f(x)$ and integrating, we get
$$
\begin{align}
\log(f(x))
&=\int\frac{2x+1}{x(1-x)}\,\mathrm{d}x\\
&=\int\left(\frac1x+\frac3{1-x}\right)\,\mathrm{d}x\\
&=\log\left(\frac x{(1-x)^3}\right)+C
\end{align}
$$
So that if $a_1=1$,
$$
\begin{align}
f(x)
&=\frac{x}{(1-x)^3}\\
&=\sum_{n=1}^\infty(-1)^{n-1}\binom{-3}{n-1}x^n\\
&=\sum_{n=1}^\infty\binom{n+1}{n-1}x^n\\
&=\sum_{n=1}^\infty\binom{n+1}{2}x^n\\
\end{align}
$$
Therefore,
$$
a_n=\binom{n+1}{2}
$$
