Taylor expansion of rational function

Question

I want to taylor expand the following function (in the complex plane):

$$(1-\frac{z^2}{3!}+\frac{z^4}{5!})^{-1}$$

I am told that the result is:

$$1+\frac{z^2}{3!}+\frac{14}{6!}z^4+O(z^6)$$

I do not see how to obtain this. The only thing I have tried is using the taylor expansion of $(1-z)^{-1}$. But using this I can only obtain the second term of the expansion written above.

Answer 1

HINT:

Recall from Taylor's Theoremthat $\frac{1}{1-x}=1+x+x^2+O(x^3)$

Now, set $x= z^2/3!-z^4/5!$ and don't forget the second order term $(z^2/3!-z^4/5!)^2$.