Radius of convergence of the series $sumlimits_{n=1}^infty (-1)^nfrac{1cdot3cdot5cdots(2n-1)}{3cdot6cdot9cdots(3n)}x^n$

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}

I feel like it's simpler than I expect but I just can't get it. How do I do this?

Answer in book: $\dfrac{3}{2}$

Answer

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 $R = \displaystyle \frac{3}{2}$.