Expand $\frac{1}{1+z+z^2}$ into a power series. $\frac{1}{1+z+z^2}=\sum_{n=0}^\infty (-1)^n (z+z^2)^n=\dots$. What is wrong with this solution?

Question

Problem:
Expand $\frac{1}{1+z+z^2}$ into a power series.

Answer:
$\frac{1}{1+z+z^2}=1-z+z^3-z^4+z^6-z^7+\dots$ and the radius of convergence is $1$.

I found a strange solution in a book:

$$\frac{1}{1+z+z^2}=\sum_{n=0}^\infty (-1)^n (z+z^2)^n\\ =\sum_{n=0}^\infty (-1)^n\left(z^n+nz^{n+1}+\frac{n(n+1)}{2}z^{n+2}+\dots+z^{2n}\right)\\ =\sum_{m=0}^\infty a_m z^m,$$ where $a_m=\sum_{k=0}^\infty (-1)^{m-k}\binom{m-k}{k}$. (Note that if $k<l$ or $l<0$, then $\binom{k}{l}=0$.)
Using the formula $\binom{k}{l}=\binom{k-1}{l}+\binom{k-1}{l-1}$,
$$a_m+a_{m-1}=\sum_{k=0}^\infty (-1)^{m-k}\binom{m-k-1}{k-1}=\sum_{l=0}^\infty (-1)^{m-1-l}\binom{m-2-l}{l}=-a_{m-2}.$$ $a_0=1,a_1=-1$.
So $a_2=0, a_3=1, a_4=-1,\dots$.
So, $\frac{1}{1+z+z^2}=1-z+z^3-z^4+z^6-z^7+\dots$.

Let $z_0:=-\frac{3}{2}$.
Then, $|z_0+z_0^2|=|-\frac{3}{2}+\frac{9}{4}|=\frac{3}{4}<1$.
So, $\sum_{n=0}^\infty (-1)^n (z_0+z_0^2)^n$ converges.
So, from the above strange solution, $1-z_0+z_0^3-z_0^4+z_0^6-z_0^7+\dots$ converges.
But $1-z_0+z_0^3-z_0^4+z_0^6-z_0^7+\dots$ does not converge because the radius of convergence is $1$.
What is wrong with the above strange solution?

Answer 1

You have two representations that are equal if they both converge. But nothing requires them both to converge for the same domain of $z$.

There are many examples of this phenomenon of different domains of convergence for nominally equivalent series. For example, the familiar Taylor series for $\ln(1+x)$ converges only when the argument of the logarithm is $\in(0,2]$, but this series can be used to derive an "overlapping" one that converges for all positive arguments.