# (easy) rearranging of power series denominator

My teacher has done this: $$\frac{1}{z^3(1-z^2/3+O(z^4))} = \frac{1+z^2/3+O(z^4)}{z^3}$$ How does that work? I don't understand why he can claim this.

-
What is $n$ here? –  Qiaochu Yuan May 30 '11 at 12:06
It's 4. I edited the expression. –  A. Top May 30 '11 at 12:11
Isn't it simply because $\frac{1}{1-u} = 1+u+O(u^2)$ ?
In case OP finds this a little opaque, the idea is to let $u=z^2/3-O(x^4)$ and notice that $u^2=O(z^4)$. –  Gerry Myerson May 30 '11 at 12:19