Are these two distinct series representations for $\frac{1}{1-x^2}$ correct? Let $f(x)=\frac{1}{1-x^2}$. 
With a $u=x^2$ substitution, $f(x)$ becomes $g(u)=\frac{1}{1-u}$, which being the limit of a geometric series has the form $$g(u)=\sum_{k=0}^\infty u^k$$
Plugging the substitution back in yields a power series representation for $f(x)$
$$\frac{1}{1-x^2}=\sum_{k=0}^\infty (x^2)^k=\sum_{k=0}^\infty x^{2k}$$
A different approach would be to factor the denominator of $f(x)$
$$\frac{1}{1-x^2}=\frac{1}{(1-x)(1+x)}=\frac{1}{1+x}\frac{1}{1-x}=\frac{1}{1+x}\sum_{k=0}^\infty x^k=\sum_{k=0}^\infty \frac{x^k}{1+x}$$
Here we have two different series representations for the same function but the second one is not a power series representation, so I'm guessing there's no issues here (unless I've made some mistake in either of my derivations).

We have
\begin{align}
\sum_{k \ge 0} \frac{x^k}{1 + x} &= \sum_{k \ge 0} \sum_{\ell \ge k} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} \sum_{k \le \ell} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} x^\ell \sum_{k \le \ell} (-1)^{\ell-k}
\end{align}
where
$$ \sum_{k \le \ell} (-1)^{\ell-k} = \begin{cases} 0 & \text{if $\ell$ is odd} \\ 1 & \text{if $\ell$ is even} \end{cases}. \tag{1} $$
In fact, the formal identity
$$ \sum_{\ell \ge 0} x^\ell \sum_{k \le \ell} (-1)^{\ell-k} = \frac{1}{1 - x^2} $$
is a great way to prove $(1)$.
