# (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.

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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

## 1 Answer

Isn't it simply because $\frac{1}{1-u} = 1+u+O(u^2)$ ?

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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
@Gerry, thanks for fleshing it out. – lhf May 30 '11 at 12:22
Thanks all! I feel so stupid. – A. Top May 30 '11 at 12:23
@A. Top: Don't! This happens to all of us :) Next time you'll remember. – t.b. May 30 '11 at 12:27