How can I find the convergence radius for this series? I want to find out the MacLaurin series of this function and find out for which $x$ it equals the original function:
$f(x)=\frac{x}{1+3x^2}$
AFAIK I can use this equation:
$(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}*x^n$.
So the MacLaurin series should be
$f(x)=x*\sum_{n=0}^{\infty}\binom{-1}{n}*(3x^2)^n$
I think the series equals the function in it's interval of convergence.
So, to find out the convergence radius, I used this approach:
$\sum_{n=0}^{\infty}a_n*(x-x_0)^n \Rightarrow r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|a_n|})^{-1}$
Then I transformed the series into the form I need:
$ \sum_{n=0}^{\infty}\binom{-1}{n}*(3^n*x^{2^n})\\
\sum_{n=0}^{\infty}(\binom{-1}{n}*3^n)*(x^{2n})\\
\sum_{n=0}^{\infty}(\binom{-1}{n}*3^n*x^{n})*(x-0)^n\\
a_n=-3^n*x^n$
And used the formula:
$
r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|a_n|})^{-1}\\
r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|-3^n*x^n|})^{-1}\\
r=(\lim\limits_{n \rightarrow \infty}\sqrt[n]{|-3^n|}*\lim\limits_{n \rightarrow \infty}\sqrt[n]{|x^n|})^{-1}\\
r=(|3|*|x|)^-1\\
r=\frac{1}{3|x|}
$
But that can't be true, can it? What is the $x$ in $r$? I can't determine the interval of convergence because I don't know what $x$ is.
 A: This is very easy to answer by knowing just a little about complex analysis and meromorphic functions. We have:
$$f(z)=\frac{z}{1+3z^2}=\frac{1}{6}\left(\frac{1}{z-\frac{i}{\sqrt{3}}}+\frac{1}{z+\frac{i}{\sqrt{3}}}\right)$$
hence $f(z)$ has two simple poles with residue $\frac{1}{6}$ at $z=\pm\frac{i}{\sqrt{3}}$ and the radius of convergence of the Taylor series of $f(z)$ around $z=0$ cannot be bigger than $\frac{1}{\sqrt{3}}$, the distance from the origin of the closest singularity. Since the radius of convergence of:
$$\sum_{n\geq 0} \alpha^n\,z^n $$
for any $\alpha\in\mathbb{C}\setminus\{0\}$ is exactly $\frac{1}{|\alpha|}$, and for every $z\in\mathbb{C}$ such that $|z|<\frac{1}{|\alpha|}$ we have:
$$\sum_{n\geq 0} \alpha^n\,z^n = \frac{1}{1-\alpha z},$$
it follows that the radius of convergence of the Taylor series of $f(z)$ around $z=0$ is exactly $\color{red}{\frac{1}{\sqrt{3}}}$.
A: The problem with your argument is that in
$$
\sum_{n=0}^{\infty}\binom{-1}{n}3^n x^n x^n
$$
doens't have Taylor coefficients $\binom{-1}{n}3^nx^n$ because the coefficients cannot be functions of $x$.
What you want to do is define $a_m$ by
$$
a_m = \cases{0 & $m$  odd \\
\binom{-1}{m/2} 3^{m/2}& $m$ even}
$$
Then we have 
$$
\sum_{n=0}^{\infty}\binom{-1}{n}3^n x^{2n} = \sum_{m=0}^\infty a_m x^m
$$
and
$$
\limsup|a_m|^{1/m} = \limsup|a_{2n}|^{1/2n} = \limsup|\binom{-1}{n}3^n|^{1/2n} = \sqrt{3}
$$
