<s>[INST] How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how to write the denominator because of the alternating signs in each term. As for the starting point, $n=?$, I think its going to be either $0$ or $1$, depending on the denominator - I don't want to be kicked from this site by attempting to divide by $0$ .
 [/INST] Once you realise that the denominators are sums of powers of $-x^2$, this becomes easy. You have
$$
  \sum_{n=0}^\infty\frac{x^n}{\sum_{i=0}^n(-x^2)^i}.
$$
Actually that denominator is a geometric series, that can be written explicitly without summation (provided that $X=-x^2\neq1$) using
$$
  \sum_{i=0}^nX^i=\frac{1-X^{n+1}}{1-X},
$$
which gives a fraction in the denominator that can be removed by simple algebra
$$
\sum_{n=0}^\infty\frac{x^n}{\sum_{i=0}^n(-x^2)^i}
=\sum_{n=0}^\infty\frac{x^n}{\frac{1-(-x^2)^{n+1}}{1-(-x^2)}}
=\sum_{n=0}^\infty\frac{x^n(1+x^2)}{1-(-x^2)^{n+1}}
=(1+x^2)\sum_{n=0}^\infty\frac{x^n}{1-(-x^2)^{n+1}}.
$$
After this I'm not sure whether it can be further simplified.
</s>