sequences and series - find $frac{3}{6}+frac{3cdot5}{6cdot9}+frac{3cdot5cdot7}{6cdot9cdot12}+cdots$

find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$

I had $(1-x)^{-\frac{p}{q}}$ in mind.
$$S=\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$

$$S+1=1+\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$
$$S+1=1+\frac{3}{2!\cdot3}+\frac{3\cdot(3+2)}{3!\cdot9}+\frac{3\cdot(3+2)\cdot(3+4)}{4!\cdot27}+\cdots$$
$$S+1=1+\frac{3}{2!}\left(\frac{\frac{2}{3}}{2}\right)+\frac{3\cdot(3+2)}{3!}\left(\frac{\frac{2}{3}}{2}\right)^2+\frac{3\cdot(3+2)\cdot(3+4)}{4!}\left(\frac{\frac{2}{3}}{2}\right)^3+\cdots$$
$$S+1=\left(1-\frac{2}{3}\right)^\frac{-3}{2}$$
I got $S=3\sqrt{3}-1$

But answer given is $S=3\sqrt{3}-4$

Answer

In the last step, you miss some multiple of $3$, and you miss one term of the expansion.

$$S=\sum_{n\geq 1} \frac{(2n+1)!!}{(n+1)!3^n}=\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\frac{(-2)^{n+1}}{3^n}=3\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\left(\frac{-2}{3}\right)^{n+1}=3\left(\left(1-\frac{2}{3}\right)^{-\frac{1}{2}}-1-\frac{1}{3}\right)=3\sqrt{3}-4.$$