Power series convergence radius My question is: how do I calculate the radius convergence of a power series when the series is not written like
$$\sum a_{n}x^{n}?$$
I have this series:
$$\sum\frac{x^{2n+1}}{(-3)^{n}}$$
Can I use the criterions as I was working with $x^{n}$, not $x^{2n+1}$? 
I tried this:
$$k=2n+1\Rightarrow n=\frac{k-1}{2}$$
And I got
$$R=\lim_{k\to\infty}\left|\frac{a_{k}}{a_{k+1}}\right|=\frac{1}{\sqrt{3}}$$
I know the answer is: the series converges for all $x$ that $|x|<\sqrt{3}$. How do I get it?
Thanks :)

Since $x\ne0$, we can apply the ratio test for Absolute Convergence, which is stated precisely as so:
Let $ \sum u_{k}$ be a series with nonzero terms and suppose that 
$$ p= {Lim_\xrightarrow{k\to\infty}}{\frac{|u_k+1|}{|u_k|}}$$
(a) The series converges absolutely if $p<1$.
(b) The series diverges if $p>1$ or $ p = \infty$
(c) The test is inconclusive if $p=1$
Applied in this scenario, we get:
$$\frac{|x^{2(n+1)+1}|}{|-3^{n+1}|}* \frac{|-3^n|}{|x^{2n+1}|}
=\frac{|x^2|}{|-3|}=\frac{x^2}{3}$$
Now $$ {Lim_\xrightarrow{n\to\infty}}{\frac{x^2}{-3}}<1\iff|x|<\sqrt3$$
Our radius of convergence is $\sqrt3$ and the interval of convergence is $-\sqrt3< x < \sqrt3 $
