Q:

Inconsistency in binomial expansion?

When I do a binomial expansion on $\frac{1}{(1-2x)(1+3x)}$ about $0$, I can do it in 2 ways.
Method 1
$$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1-2x}\frac{1}{1+3x}$$
Thus, getting $(1+y_1)^{-1}(1+y_2)^{-1}$.
This method would give me a radius of convergence $|2x|<1$ AND $|3x|<1$. So, the radius of convergence is $-\frac{1}{3}< x < \frac{1}{3}$.
Method 2
$$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1+x-6x^2}$$
I can let $y=x-6x^2$, thus getting $(1+y)^{-1}$.
Now, the convergence will be for $|y|<1$, or $|x-6x^2|<1$.
This actually gives me $-\frac{1}{3} < x < \frac{1}{2}$.
So, I have $\frac{1}{3}<x<\frac{1}{2}$ in my convergence as well.
Question: So which is the correct method and why is there such an inconsistency?
I have my own theories, but I'll see what everyone has to say before weighing in.

A:

Assuming you are doing the expansion around $0$, you get convergence out to the nearest root (in the complex plane).  In both cases this is $\frac 13$.  The error is assuming that the radius of convergence of $\frac 1{1+x-6x^2}$ is that with three terms in the denominator, they pollute each other in the power series.  They do it in just the way to make Method 1 correct.

