Radius of convergence of the series $ \sum\limits_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{3\cdot6\cdot9\cdots(3n)}x^n$ How do I find the radius of convergence for this series:
$$ \sum_{n=1}^\infty (-1)^n\dfrac{1\cdot3\cdot5\cdot\cdot\cdot(2n-1)}{3\cdot6\cdot9\cdot\cdot\cdot(3n)}x^n $$
Treating it as an alternating series, I got 
$$x< \dfrac{n+1}{2n+1}$$
And absolute convergence tests yield
$$-\dfrac{1}{2}<x<\dfrac{1}{2}$$
I feel like it's simpler than I expect but I just can't get it. How do I do this?
Answer in book: $\dfrac{3}{2}$

The ratio test allows to determine the radius of convergence.
For $n \in \mathbb{N}^{\ast}$, let :
$$ a_{n} = (-1)^{n}\frac{1 \times 3 \times \ldots \times (2n-1)}{3 \times 6 \times \ldots \times (3n)}. $$
Then, 
$$ \begin{align*}
\frac{\vert a_{n+1} \vert}{\vert a_{n} \vert} &= {} \frac{1 \times 3 \times \ldots \times (2n-1) \times (2n+1)}{3 \times 6 \times \ldots (3n) \times (3n+3)} \times \frac{3 \times 6 \times \ldots \times (3n)}{1 \times 3 \times \ldots \times (2n-1)} \\[2mm]
&= \frac{2n+1}{3n+3} \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; \frac{2}{3}.
\end{align*}
$$
Since the ratio $\displaystyle \frac{\vert a_{n+1} \vert}{\vert a_{n} \vert}$ converges to $\displaystyle \frac{2}{3}$, we can conclude that $R = \displaystyle \frac{3}{2}$.
