What values of x does $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converge for? In a past year paper there was a question asking to find the values of x in which  $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converges.
I tried using the root test, and so:
$\lim_{n \rightarrow \infty} (\frac{1}{n} (\frac{3}{(-1)^n+2}x^{n})^{1/n} = \lim_{n\rightarrow \infty} \frac{1}{n^{1/n}} \frac{3}{-1+2^{1/n}} x$
And so we have a problem because $lim_{n \rightarrow \infty} 2^{1/n} = 1$ and we will get 0 in the denominator. 
Help would be much appreciated!

First note that the coefficients of this power series is
$$a_n=\frac{1}{n}\left(\frac{3}{(-1)^n+2}\right)^n=\begin{cases}\frac{3^n}{n},&n\text{ odd}\\
\frac{1}{n},&n\text{ even}\end{cases}.$$
Then for sufficiently large $n$
$$\frac{3}{n^\frac{1}{n}}\leq \sup_{N\geq n}|a_N|^{\frac{1}{N}}\leq 3.$$
Thus the radius of convergence is
$$R=\frac{1}{\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}}=\frac{1}{3}.$$
Now we consider the case $|x|=\frac{1}{3}$. It's obvious that the series diverges when $x=\frac{1}{3}$. For $x=-\frac{1}{3}$, then the odd terms of the series are $-\frac{1}{n}$ and the even terms are $\frac{1}{n3^n}$, and the series also diverges in this case.
All in all, the series converges precisely when $|x|<\frac{1}{3}$.
