<s>[INST] 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?
 [/INST] 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.
</s>