Rewriting a Binomial expansion and expanding the rewritten version after making a convenience substitution to get a totally different answer I know that if I have a function $$f(x) = \frac{1}{x-1}$$ that if I perform a binomial expansion on I get the  result $$f(x)\approx -1 -x -x^2-x^3-x^4 ...$$
However if I rewrite the function $$f(x) = \frac{1}{x-1} = \frac{1}{x(1-\frac{1}{x})} = \bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$$
Then we make the expansion of $\bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$ by using the subsitution $\frac{1}{x} = z$
We get $$(1-z)^{-1} = 1+z+z^2+z^3+z^4... = 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+ \frac{1}{x^4}...$$
so that
$$x^{-1}(1-z)^{-1} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4}+ \frac{1}{x^5}...$$
This is completely different result!
Why is this the case?

The expansion of $\frac 1 {x-1}$ you are using is valid only for $|x|<1$. When you change $x$ to $\frac 1 x$ and use the expansion you are assumung that $|x|>1$. Since there is no number with $|x|<1$ and $|x| >1$ you cannot use both of them for the same $x$.
