Series with $n$th term having integer raised to the power of $n$ in the denominator

$$1+\frac{4}{6}+\frac{4\cdot5}{6\cdot9}+\frac{4\cdot5\cdot6}{6\cdot9\cdot12}+\cdots$$

I could reduce it to $n$th term being $\dfrac{(n+1)\cdot(n+2)}{n!\cdot3^n}$.
Took me an hour just to get to this.
But I am now stuck up. PL. Help

Answer

$$1+\frac{4}{6}+\frac{4\cdot5}{6\cdot9}+\frac{4\cdot5\cdot6}{6\cdot9\cdot12}+\cdots$$
$$=1+\sum\limits_{n=1}^\infty\frac{(n+3)!}{2\times3\times3^n\times(n+1)!}$$
$$=1+\frac{1}{2}\sum\limits_{n=1}^\infty\frac{(n+2)\times(n+3)}{3^{n+1}}$$

By induction, we can show that for $n\ge7$, $0<\frac{(n+2)\times(n+3)}{3^{n+1}}<\frac{1}{n^2}$, and hence the series is convergent.

In fact, $=1+\frac{1}{2}\cdot\frac{11}{4}=\frac{19}{8}$