Correct way of determining the range of validity when expanding (1+f(x))^n Consider the following three ways to expand $\frac{1}{1+3x+2x^2}$ = $\frac{1}{(1+x)(1+2x)}$ = $\frac{2}{1+2x}$ $-$ $\frac{1}{1+x}$ using the standard binomial series expansion for $(1+x)^n$ valid for $|x|<1$:
(a) $(1+y)^{-1}$ where $y=3x+2x^2$ valid for $|y|<1$
(b) $(1+x)^{-1}$ * $(1+2x)^ {-1}$ valid for $|x|<1/2$
(c) $2(1+2x)^{-1}$ $-$ $(1+x)^{-1}$ valid for $|x|<1/2$
These methods all seem to give the correct power series as far as you want to go.
Now with (b) and (c) the correct range of validity, $-0.5<x<0.5$ drops out clearly.
However, for (a) the range of validity is $|y|<1$ which gives a different set of values for the range of validity for $x$.
What is the resolution? Abandon method (a)? Why doesn't it work?
In general, is the expansion for $(1+f(x))^n$ valid for $|f(x)|<1$?
 A: Edit: my answer is incomplete/ possibly wrong, so don't read too much into it at the moment. I think I've got a better answer coming along shortly...
"In general, is the expansion for $(1+f(x))^n$ valid for $|f(x)|<1$?"
Yes. See:
https://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence
In particular, quoting the general Binomial series from the above link:

\begin{align} (1 + x)^\alpha  &=\sum_{k=0}^{\infty} \; {\alpha
\choose k} \; x^k   \qquad\qquad\qquad (1) \\ &= 1 + \alpha x +
\frac{\alpha(\alpha-1)}{2!} x^2 + \cdots \end{align}
Whether (1) converges depends on the values of the complex numbers $\alpha$
and $x$. More precisely:
If $|x| < 1$, the series converges (absolutely) for any complex number $\alpha$.

So to clarify, seeing as you probably only care about convergence in $\mathbb{R}$ rather than $\mathbb{C}$, if there is a domain $D = \{{s:|f(s)|< 1}\} \subseteq \mathbb{R}$ then the (general) Binomial expansion of $(1+f(x))^n$ converges on D for any real number $n$. (Note: absolute convergence implies convergence).
If it's not clear what I'm getting at, think about it like this: The (Newton's) general Binomial expansion is just saying that, for some numbers, $(1) $ works. What I'm saying in the previous paragraph is, for any number $a \in D$, $f(a)$ is also just a number for which $(1)$ works.
For other cases e.g. when $|f(x)| = 1$, see the link above and just apply it to $\mathbb{R}$, which is fine because obviously $\mathbb{R} \subseteq\mathbb{C}$
Also, see the wikipedia link for proofs (e.g. proof of why the Binomial expansion works).
Dislaimer: The rest of my answer is incomplete, and could probably be revised by someone more familiar with the ins and outs of conditional convergence / Riemann Series Theorem, of which I only have a baseline knowledge.
Back to your examples (a), (b) and (c). Why do they have different radii of convergence?
Well, my guess would be that (b) and (c) tell you that  $|x| < \frac{1}{2} \implies $ the binomial expansion for those functions converge. But that does not mean they do not converge outside of $|x| < \frac{1}{2}$: it does not say it is valid for only $|x| < \frac{1}{2}$. Here, you've got two alternating series, so assuming they are conditionally convergent (are they?), then by the Riemann Series Theorem, a rearrangement of terms can result in any real number.
See: https://en.wikipedia.org/wiki/Riemann_series_theorem
I'm not sure about (b) because I'm not sure about multiplying two alternating series- it might also depend how you multiply the two brackets.
But for series (c) my guess would be that
$ \qquad 2(1+2x)^{-1} - (1+x)^{-1}
\\
 \quad =2(1-2x+4x^2 +...) - (1-x+x^2+...)$
and if you do not re-arrange terms, this indeed converges for only $|x| < \frac{1}{2}$.
But if you rearranged and grouped the x-coefficients to get:
$ \qquad 2(1+2x)^{-1} - (1+x)^{-1}
\qquad= 2(1-2x+4x^2 +...) - (1-x+x^2+...)
\qquad= 1-3x + 7x^2 -15x^3+...$,
which is what I imagine the expansion of (a) is,
then maybe the radius of convergence really is different, and this is permitted by the Riemann Series Theorem. i.e. different re-arrangements of the series will give different values of x converging (and diverging).
