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 $\frac{11}{576x^6}$.)

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 $\color{red}{a_6=-\frac{19}{576}}$.
