How do I simplify $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ using long division? The infinite series $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ is supposed to simplify to $1-\frac{x^2}{2}+\frac{7x^4}{48}+\frac{19x^6}{576}\cdots$ but I don't know how this was calculated.
Did I make an error or did Zill make an error on the $x^6$ term (I got $\frac{11}{576x^6}$.)
 A: Actually, I got
$$1-\frac{x^2}{2}+\frac{7x^4}{48}-\frac{19x^6}{576}\cdots$$
Let
$$D = 1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots,$$
your divisor.
First remainder is
$$R_1 = 1 - 1D = -\frac{x^2}{2} - \frac{5x^4}{48} - \frac{7x^6}{576} + O(x^8).$$
Second remainder (for the $-x^2/2$ term) is 
$$R_2 = R_1 - \left(-\frac{x^2}{2}\right)D = \frac{7x^4}{48} + \frac{23x^6}{576} + O(x^8).$$
Third remainder (for the $5x^4/48$ term) is
$$R_3 = R_2 - \left(\frac{7x^4}{48}\right)D = -\frac{19x^6}{576} + O(x^8),$$
which gives the last term.
A: Assume that:
$$\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots} = 1+a_2 x^2+a_4 x^4 + a_6 x^6 + \ldots $$
and multiply both sides by $1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots$. If, after that, we consider the coefficient of $x^2$ in both sides, we get $\frac{1}{2}+a_2=0$, so $a_2=-\frac{1}{2}$. If we consider the coefficient of $x^4$, we get $\frac{5}{48}+\frac{1}{2}a_2+a_4=0$, hence $a_4=\frac{7}{48}$. If we consider the coefficient of $x^6$, we get:
$$ \frac{7}{576}+\frac{5}{48}a_2 + \frac{1}{2}a_4 + a_6 = 0, $$
hence $\color{red}{a_6=-\frac{19}{576}}$.
