Landau notation on $f(n) = \frac{1}{4n \tan(\frac{\pi}{n})}$ Does someone would have time to show me how to use the Landau notation "Big O"? A useful example could be on $f(n) = \frac{1}{4n \tan(\frac{\pi}{n})}$.
 A: $f(x)=O(x)$ as e.g. $x\to \infty$ means that there exists a constant $M>0$ such that $|f(x)|\leq Mx $ for sufficiently large $x$.
Lets analyze your example. First note that the powerseries of cotangent is
$\cot x= \frac{1}{\tan x}= \frac{1}{x}-\frac{1}{3}x -\dots$  where the rest of the terms involve powers $x^m, m\geq 3$. Therefore, for $x$ close to zero these terms are very small comparing to $x$, so you can write $\cot x= \frac{1}{x}- \frac{1}{3}x + O(x^3)$. This means that there exists a constant $M>0$ such that
$$|\cot x-(\frac{1}{x}- \frac{1}{3}x)| \leq M|x|^3$$
for sufficiently small $x$. 
In your case $f(n)=\frac{1}{4n\tan(\pi/n)}=\frac{1}{4n}\cot(\pi/n)= \frac{1}{4\pi}- \frac{\pi}{12n^2}+O(1/n^4)$
where we have an extra power in the $O$ coming from the term $1/4n$. 
A: One has, by the Taylor series expansion, as $x \to 0$,
$$
\frac{1}{1+x}=1+O(x),\qquad \tan x=x+O(x^3),\tag1
$$ then
$$
\frac{x}{\tan x}=\frac{x}{x+O(x^3)}=\frac1{1+O(x^2)}=1+O(x^2)\tag2
$$ Hence, as $n \to \infty$, we have $\dfrac{\pi}n \to 0$, and
$$
\begin{align}
f(n) := \frac{1}{4n \tan(\frac{\pi}{n})}
=\frac1{4\pi}\cdot \frac{\frac{\pi}{n}}{\tan(\frac{\pi}{n})}
=\frac1{4\pi}\cdot \left( 1+O\left(\frac1{n^2}\right) \right)
=\frac1{4\pi}+O\left(\frac1{n^2}\right) \tag3
\end{align}
$$ as wanted.
