Proving an identity using formal power series 
4.
(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.
(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?
 A: For the first, $$ \frac{1}{1 + \frac{x^2}{1 - x}} = \frac{1 - x}{1 - x + x^2} = \frac{(1 + x)(1 - x)}{(1 + x)\left(1 - x + x^2\right)} = \frac{1 - x^2}{1 + x^3} $$
For the second, consider the Maclaurin Series of $ \frac{1 - x^2}{1 + x^3} $ which is $$ \left( 1 - x^2\right)\sum_{k = 0}^\infty \left(-x^3\right)^n = \sum_{k = 0}^\infty a_nx^n$$ where $$ a_n = \begin{cases} &0 & x \equiv 1 \mod 3 \\ -&1 & x \equiv 2 \mod 6 \vee x \equiv 3 \mod 6 \\ &1 &x \equiv 5 \mod 6 \vee x \equiv 0 \mod 6 \end{cases} $$
Meanwhile, he Maclaurin Series for $ \frac{1}{1 + \frac{x^2}{1 - x}} $ is $ \displaystyle \sum_{n = 0}^\infty f(n) x^n $ where $$ f(n) := \sum_{k\ge 0 } (-1)^k  \dbinom{n -k -1}{n - 2k} $$Because the two functions are equivalent near $ x = 0 $, their Maclaurin series are the same and the equality follows.
The hard part is proving that definition of $ f(n) $. It involves expanding $ \frac{x^2}{1 - x} $ as its own Maclaurin series and then using a geometric series on $ \frac{1}{1 - \left(-\frac{x^2}{1 - x}\right)}.
A: As DonAntonio points out, the first part is just a matter of factoring and doing some algebra. For the second part, note first that
$$\begin{align*}
\frac{1-x^2}{1+x^3}&=(1-x^2)\cdot\frac1{1-(-x^3)}\\
&=(1-x^2)\sum_{n\ge 0}(-1)^nx^{3n}\\
&=\sum_{n\ge 0}(-1)^nx^{3n}-\sum_{n\ge 0}(-1)^nx^{3n+2}\;,
\end{align*}$$
in which the coefficient of $x^{3n}$ is $(-1)^n$, the coefficient of $x^{3n+1}$ is $0$, and the coefficient of $x^{3n+2}$ is $(-1)^n$. Clearly, then, the absolute value of the coefficient of $x^N$ is $0$ if $N\equiv 1\pmod 3$ and $1$ otherwise.
Similarly,
$$\begin{align*}
\frac{1}{1+\frac{x^2}{1-x}}&=\frac1{1-\left(-\frac{x^2}{1-x}\right)}\\
&=\sum_{n\ge 0}(-1)^n\left(\frac{x^2}{1-x}\right)^n\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\frac1{(1-x)^n}\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\sum_{k\ge 0}\binom{n+k-1}kx^k\;,\tag{1}
\end{align*}$$
where the last step uses the standard power series expansion of $\dfrac1{(1-x)^n}$.
The coefficient of $x^N$ in $(1)$ is
$$\sum_{n\ge 0}(-1)^n\binom{N-n-1}{N-2n}\;;$$
to get this I simply write $N=2n+k$ and sum over $n$, so that $k=N-2n$. (The binomial coefficient is $0$ when $n$ is too large, so there’s no need to specify an upper bound.) 
The result is now immediate.
A: Hints:
$$1-x^2=(1-x)(1+x)$$
$$1+x^3=(1+x)(1-x+x^2)\;\;\ldots$$
A: To your question, “How should I expand a fraction into a power series?”, the hand-computational answer is this. You write numerator and denominator down with powers ascending to the right and then proceed exactly as you did in high school when you divided one polynomial into another.
