How is $\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)^{-1}$ equal to $\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)$?

Question

In the book of Concrete Mathematics by Graham, Knuth, Patashnik, at page 453, it is given that

$$ \begin{aligned} \lfloor\mathrm{N} / \mathrm{K}\rfloor &=\mathrm{N}^{1-1 / 3}\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)^{-1}+\mathrm{O}(1) \\ &=\mathrm{N}^{2 / 3}\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)+\mathrm{O}(1)=\mathrm{N}^{2 / 3}+\mathrm{O}\left(\mathrm{N}^{1 / 3}\right) \end{aligned} $$ where $$ \mathrm{K}=\mathrm{N}^{1 / 3}\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right). $$

However I don't understand how $\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)^{-1}$ is equal to $\left(1+\mathrm{O}\left(\mathrm{N}^{-1 / 3}\right)\right)$ as done above.

Answer 1

From Taylor formula/series for $\frac{1}{1+x}$ we have $$\frac{1}{1+x} = 1+O(x)\implies (1+O(N^{-1/3}))^{-1} = 1+O(O(N^{-1/3})) = 1+O(N^{-1/3}).$$

Answer 2

Suppose $f(N)=1+O(N^{-1/3})$, which means there exists a constant $C>0$ and integer $M>0$ such that for each $N>M$ we have $|f(N)-1|<C\cdot N^{-1/3}$. Note that since $f(N)>1-C\cdot N^{-1/3}$, for large enough $N$ ($>8C^3$) we have $f(N)>\frac12$.

Now, we wish to prove $f(N)^{-1}=1+O(N^{1/3})$. We have, for $N$ large enough ($N>\max\{M,8C^3\}$), that $|f(N)^{-1}-1|=|f(N)|^{-1}\cdot |f(N)-1|<2\cdot (C\cdot N^{-1/3})$. Now, taking $2C$ as our new constant, we have shown $f(N)^{-1}=1+O(N^{1/3})$, as desired.