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 $0$ .
 A: The denominators alternate signs, but they are of the form
$$\sum_{k=0}^m (-1)^k x^{2 k} = \frac{(-1)^{m+1} x^{2(m+1)}-1}{x^2+1}$$
The sum may then be written as
$$(x^2+1) \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(-1)^{n+1} x^{2(n+1)}-1} = -(x^2+1) \sum_{n=0}^{\infty} \frac{x^n}{x^{2(n+1)}+(-1)^n}$$
I should note that convergence is achieved when $|x|<1$ when $x$ is real.
A: HINT
The denominators in every term, except the first, are the alternating sums of even powers.
\begin{array}
.x^2-1 &\equiv& x^2 - x^0 \\
x^4-x^2+1 &\equiv& x^4 - x^2 + x^0 \\
x^6-x^4+x^2-1 &\equiv& x^6-x^4+x^2-x^0
\end{array}
Putting these in to summation notation:
\begin{array}
.x^2-x^0 &\equiv& \sum_{k=0}^1 (-1)^{k}x^{2(1-k)} \\ \\
x^4-x^2+x^0 &\equiv& \sum_{k=0}^2 (-1)^{k}x^{2(2-k)} \\ \\
x^6-x^4+x^2-x^0 &\equiv& \sum_{k=0}^3 (-1)^{k}x^{2(3-k)} \\ \\
\cdots\cdots\cdots &\equiv& \sum_{r=1}^n\sum_{k=0}^r(-1)^kx^{2(r-k)}
\end{array}
A: 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.
