How I can prove equality below?
$$ \frac{1}{1 + O(n^{-1})} = 1 + O({n^{-1}}), $$ where $n \in \mathbb{N}$ and we are considering situation when $n \to \infty$.
It is clearly that it is true. But I don't know which property I have to use to prove it formally. I will appreciate if someone could give me some clues.
On
If $n \rightarrow \infty$ then $O(\frac{1}{n}) \rightarrow 0$ and using $$\frac{1}{1-z} = \sum_{n=0}^{\infty}z^{n} \ \ \ \ , \ \ |z|<1$$ we have $$\frac{1}{1 + O(n^{-1})} = \sum_{k=0}^{\infty}(-1)^{k}O(\frac{1}{n})^{k} = 1 +O(\frac{1}{n})$$
On
Assume $f\in O(n^{-1})$. Then there exists $c\in\mathbb R$ and $n_0\in\mathbb N$ such that $|f(n)|<\frac cn$ for all $n>n_0$. Wlog. $n_0>2c$. Then for $n>n_0$ we have $1+f(n)>1-\frac cn>1-\frac c{n_0}>\frac12$ and hence $$ 0<\frac1{1+f(n)}<\frac1{1-\frac cn}=1-\frac{c}{n-c}$$ and from $-\frac c{n-c}\in O(n^{-1})$ the claim.
We know that $$ f(n)=1+{\mathcal O}(n^{-1}), $$ iff $$ |f(n)-1|\le cn^{-1}, $$ for some $c>0$ and for $n\ge n\ge N$. Thus $$ 1-cn^{-1}\le f(n)\le 1+cn^{-1}, $$ for $n\ge N$ and hence for $n\ge N$: $$ 1-\frac{c}{n+c}=\frac{n}{n+c}=\frac{1}{1+cn^{-1}}\le \frac{1}{f(n)}\le \frac{1}{1-cn^{-1}}=\frac{n}{n-c}=1+\frac{c}{n-c}, $$ and hence $$ \left|\frac{1}{f(n)}-1\,\right|\le \max\left\{\frac{c}{n-c},\frac{c}{n+c}\right\}=\frac{c}{n-c}\le \frac{2c}{n}, $$ if $n\ge \max\{N,2c\}$.
Thus $1/f=1+{\mathcal O}(n^{-1})$.