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 :)
 A: Let 
$$u_n(z)=\frac{x^{2n+1}}{(-3)^n}$$
then by the ratio test we have 
$$\frac{|u_{n+1}(z)|}{|u_n(z)|}=\frac{|x|^2}{3}\xrightarrow{n\to\infty}\frac{|x|^2}{3}<1\iff|x|<\sqrt3$$
hence the radius of convergence is $R=\sqrt3$.
A: You can still use the ratio test; suppose $x\ne0$; then the ratio to compute is
$$
\frac{|x|^{2n+3}}{3^{n+1}}\bigg/\frac{|x|^{2n+1}}{3^{n}}=
\frac{|x|^{2n+3}}{3^{n+1}}\frac{3^{n}}{|x|^{2n+1}}=\frac{|x|^2}{3}
$$
This is constant, so the series is actually a geometric series, easy to analyze; but, also in general, you know that, as long as the limit of the ratios is less than $1$, the series is convergent.
This is not always applicable, but if the limit exists, you can conclude.
For a different example, consider
$$
\sum_{n\ge0}\frac{x^{2n+1}}{(2n+1)!}
$$
where the ratio to compute is
$$
\frac{|x|^{2n+3}}{(2n+3)!}\bigg/\frac{|x|^{2n+1}}{(2n+1)!}=
\frac{|x|^{2n+3}}{(2n+3)!}\frac{(2n+1)!}{|x|^{2n+1}}=
\frac{|x|^2}{(2n+3)(2n+2)}
$$
Since the limit of this as $n\to\infty$ is zero, we know that the series converges for all $x$.
A: It's better to use this:
$\rho=\limsup_{n \to \infty} \sqrt[n]{|a_{n}|}$, then the series converges for all $x$ that $|x| < \frac{1}{\rho}$ and doesn't coverge for $|x|>\frac{1}{\rho}$. In this case:
$\limsup_{n \to \infty} \sqrt[n]{|a_n|}=\limsup_{n \to \infty} \sqrt[2n+1]{\frac{1}{3}}=\frac{1}{\sqrt{3}}$, so the series converges for all $x$ that $|x| < \sqrt{3}$ and doesn't coverge for $|x|>\sqrt{3}$. For $|x|=\sqrt{3}$ the series doesn't coverge, because $a_n=- \sqrt{3}$ or $a_n=\sqrt{3}$ for all $n$.
A: 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 $
