Maclaurin series of $(1+x^3)/(1+x^2)$ I can't seem to figure out the Maclaurin series of $(1+x^3)/(1+x^2)$
I started with $ 1/(1-x)= \sum_{n=1}^{\infty} x^n  $
$ 1/(1-(-x^2)) = \sum_{n=1}^{\infty}(-1)^n x^{2n}  $
$ (1+x^3) \sum_{n=1}^{\infty}(-1)^n x^{2n}= [\sum_{n=1}^{\infty}(-1)^n x^{2n}]+x^3 [\sum_{n=1}^{\infty}(-1)^n x^{2n}] $
$ [\sum_{n=1}^{\infty}(-1)^n x^{2n}]+[\sum_{n=1}^{\infty}(-1)^n x^{2n+3}] $  ....
Here I'm stuck. I don't know how to merge the two sums. Maybe I have made a mistake. When I write out the sum I get:
$ 1 -x^2 +x^3 + x^4 - x^5 - x^6 + x^7 + x^8 -x^9 - x^{10} +...   $
So it seems that you need to write out a 'simple' $\sum_{n=1}^{\infty} x^n $ but I don't know how to do the start and how to distribute the minuses. Or maybe I made a mistake.
 A: Don't do more work than you need to.  The better approach is to perform the polynomial long division first:
$$\frac{1+x^3}{1+x^2} = x + \frac{1-x}{1+x^2}.$$  Now apply your series formula:
$$\begin{align}
x + \frac{1-x}{1+x^2}
&= x + \sum_{n=0}^\infty (1-x)(-x^2)^n \\
&= x + \sum_{n=0}^\infty (-1)^n (x^{2n} - x^{2n+1}).\\
\end{align}$$
You could leave this as is, or you could observe that the $x$ outside the sum cancels with a term in the sum:  when $n = 0$, the summand is $1 - x$, thus an alternative form is
$$\frac{1+x^3}{1+x^2} = 1 + \sum_{n=1}^\infty (-1)^n (x^{2n} - x^{2n+1}). \tag{1}$$
But this is still somewhat unsatisfactory, since what would be ideal is to compute $f^{(n)}(0)$ in terms of $n$; i.e., the Maclaurin expansion has the form $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n.$$  Using $(1)$ to write out the first few terms,
$$\frac{1+x^3}{1+x^2} = 1 - x^2 + x^3 + x^4 - x^5 - x^6 + x^7 + x^8 - \cdots,$$ so this suggests $|f^{(n)}(0)| = n!$ except when $n = 1$, and the sign is positive if $n$ is congruent to $0$ or $3$ modulo $4$; i.e., $$f^{(n)}(0) = \begin{cases} 0 & n = 1, \\ n! & n \in \{4m, 4m+3\}, \\ -n! & n \in \{4m+1, 4m+2\} \cap n \ne 1. \end{cases}$$
A: The answer you are conjecturing can be written as
$$\frac{1+x^3}{1+x^2}=1-x^2+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})$$
Multiplying both sides by $1+x^2$ gives us
\begin{align*}
1+x^3&=(1+x^2)(1-x^2)+(1+x^2)\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(1+x^2)(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})
\end{align*}
The sum on the RHS is telescoping; that is, $\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})$ is equal to
\begin{align*}
(x^3+x^4+x^5+x^6)-(x^5+x^6+x^7+x^8)+(x^7+x^8+x^9+x^{10})-\dots=x^3+x^4.
\end{align*}
Therefore,
$$1+x^3=1-x^4+x^3+x^4=1+x^3.$$
You should be able to work backwards to fomulate a proof.
A: $$\frac{1+x^3}{1+x^2} =x+\frac{1-x}{1+x^2}=x+\frac{1}{1+x^2}-\frac{x}{1+x^2}$$
So you can use the Maclaurin series for $\frac{1}{1+x^2}$ to get the answer.
