Using Taylor series with remainder

Question

Arfken and Weber expand $(1+n^{-2})^{-1}$ as $$(1+n^{-2})^{-1}=1-n^{-2}+n^{-4}-\frac{n^{-6}}{1+n^{-2}}$$ However, if I use Taylor/Maclaurin expansion $$(1+x)^m=1+mx+ \frac{m(m-1)}{2!} x^2 + \cdots + R_k$$ with $m=-1$, $x=n^{-2}$ and remainder, $$R_k=\frac {x^k}{k!}(1+x)^{m-k} m(m-1) \cdots (m-k+1)$$ I get $$(1+n^{-2})^{-1}=1-n^{-2}+n^{-4}-\frac{n^{-6}}{(1+n^{-2})^{4}}$$ Note that the only difference is in the denominator of last term (remainder). So where did I go wrong?

Answer 1

What they did is this

$$ (1+x)^{-1} = 1-x+x^2-x^3( 1 - x + x^2 -\dots ) $$

$$ = 1-x+x^2 -\frac{x^3}{1+x}. $$

Now subs $x=1/n^2$.