Parabolic asymptote of $n\cot\frac\pi{2n}$

Question

I have determined that $$\lim_{n\to\infty}\frac{n\cot\frac\pi{2n}}{n^2}=\lim_{n\to0^+}n\cot{\frac{\pi n}2}=\lim_{n\to0^+}\frac{n\cos{\frac{\pi n}2}}{\sin{\frac{\pi n}2}}=\frac2\pi$$

So that the function $n\cot\frac\pi{2n}$ is asymptotic to $\frac2\pi n^2$. However, to my surprise, it appears that the function actually has a parabolic asymptote; in fact, it appears that $$\lim_{n\to\infty}\left(n\cot\frac\pi{2n}-\frac2\pi n^2 \right)=-\frac\pi6$$

This implies that $n\cot\frac\pi{2n}$ has a parabolic asymptote of $\frac2\pi n^2-\frac\pi6$. So my question is how do we show that the above equation is true? Also, is there any reason why the function $n\cot\frac\pi{2n}$ should have a parabolic asymptote? Is it not something more conventional of rational functions?

Answer 1

Hint

For small values of $x$ $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ Replace $x$ by $\frac{\pi}{2n}$ to get $$\cot\Big(\frac{\pi}{2n}\Big)=\frac{2 n}{\pi }-\frac{\pi }{6 n}-\frac{\pi ^3}{360 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

I am sure that you can take from here.