$\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence

Question

Could anyone is able to give me a hint why $\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence?

Answer 1

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 $t\in\mathbb{R}$.

Answer 2

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 $n\geq 1$.

Answer 3

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 $n\geq 3$.