How to find radius of convergence of given series? Consider the series
$$\frac{1}{3}+\frac{x}{5}+\frac{x^2}{3^2}+\frac{x^3}{5^2}+\frac{x^4}{3^3}+\frac{x^5}{5^3}+\cdots$$
I want to find the radius of convergence of this series.
$\textbf{Here is my work}$
I tried to figure out $a_n$ for the given power series but it did not work. 
$$a_n=3^{-\frac{n+1}{2}|\sin(\frac{n\pi}{2})|} \times5^{-\frac{n}{2}|\sin(\frac{(n+1)\pi}{2})|}$$
$$L=\lim_{n\to \infty}\frac{a_{n+1}}{a_n
}
$$
Also I thought about finding radius of convergence separately for even and odd terms but I am not sure how is that going to help.

The series:
$$
S=\frac{1}{3}+\frac{x}{5}+\frac{x^2}{3^2}+\frac{x^3}{5^2}+\frac{x^4}{3^3}+\frac{x^5}{5^3}+...
$$
Can be expressed as:
$$
S=\sum_{k=1}^{\infty} {a_k}
$$
Where:
$$
a_k=\frac{x^{k-1}}{3^k}+\frac{x^{k}}{5^k}
$$
Now, testing the convergency:
$$
\lim_{k \rightarrow \infty}\left|{\frac{a_{k+1}}{a_k}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{\frac{x^{k}}{3^{k+1}}+\frac{x^{k+1}}{5^{k+1}}}{\frac{x^{k-1}}{3^k}+\frac{x^{k}}{5^k}}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{\frac{1}{3^{k+1}}+\frac{x}{5^{k+1}}}{\frac{1}{3^k x}+\frac{1}{5^k}}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{1+x\left(\frac{3}{5}\right)^{k+1}}{\frac{3}{x}+3\left(\frac{3}{5}\right)^k}}\right|=...
$$
$$
...=\lim_{k \rightarrow \infty}\left|\frac{x}{3}\right|<1
$$
Hence, we get:
$$
\left|x\right|<3
$$
And finally radius of convergence is $r=3$.
