$\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence Could anyone is able to give me a hint why $\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence?
 A: One may observe that
$$
\left(\frac{x}{\tan x} \right)'=\frac{\sin 2x-2x}{2\sin^2 x}<0, \quad 0<x<1,
$$ giving that $x \mapsto \dfrac{x}{\tan x} $ is decreasing, then since $n \mapsto \dfrac{\pi}n$ is decreasing we have that $n \mapsto \dfrac1{4\pi} \cdot\dfrac{x}{\tan x}\circ\dfrac{\pi}n$ increasing for $n\geq 1$.
A: Function
$$f(x)=\frac{1}{4x\tan{\frac{\pi}{x}}}$$ is increasing, since $f'(x)>0$, which is easily proved:
$$f'(x)=\frac{-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}}{4x^2\tan^2{\frac{\pi}{x}}}>0$$ 
is equivalent with $$-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}>0,$$
which is again equivalent with inequality
$$-4\sin{\frac{\pi}{x}}\cos{\frac{\pi}{x}}+4\frac{\pi}{x}>0,$$
that holds, since $\sin{t}<t$ and $\cos{t}<1$ for every $t\in\mathbb{R}$.
A: Since $f(x)=\tan(x)$ is an odd function and a solution of the DE
$$ f'(x)=1+f(x)^2 \tag{1}$$
we have that $\tan(x)$ is a strongly monotonic function, i.e.
$$ \tan(x) = c_1 x+c_3 x^3+c_5 x^5+ c_7 x^7+\ldots\tag{2} $$
for any $|x|<\frac{\pi}{2}$, with $c_1,c_3,c_5,c_7,\ldots > 0$. It follows that both $\tan\frac{\pi}{n}$ and $4n\tan\frac{\pi}{n}$ give monotonic sequences for $n\geq 3$.
