(If you're good at it, you can do it by heart.)
Let $a_n$ be the average value of lengths of all the diagonal lines of regular n-gon whose side is 1.
Find a value of
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}$$
Source: Fujino_Yusui
After messing around with the cosine theorem and doing some calculations, I found the exact formula of $a_{n},$ so I did the rest of the calculations and the answer is
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}= \lim_{n\rightarrow\infty}\frac{2\cos\frac{\pi}{n}- 1}{\left ( \left ( 1- \cos\frac{\pi}{n} \right )n \right )\left ( n- 3 \right )}= \frac{2}{\pi^{2}}$$
But I want to see a shortcut that leads $a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$ without cosine theorem, I need to the help.
Best Answer
Inscribe the n-gon in a circle. When calculating the length of a diagonal, instead of using the cosine theorem, you can bisect the central angle subtended on the diagonal. The bisector, the radia leading to the end points of the diagonal, and the diagonal will create two right triangles, from which you can get that the length of the diagonal is $$ L = 2R \sin\frac{\alpha}{2}$$ where $R$ is the radius of the circle cicrumscribed on the polygon and $\alpha$ is the central angle subtended on the diagonal. You can also use this formula to calculate the length of a side of the n-gon: $$ a = 2R \sin\frac{\pi}{n}$$ Since $a=1$, we get $$ R = \frac{1}{2 \sin\frac{\pi}{n}} $$ $$ L = \frac{\sin\frac{\alpha}{2}}{\sin\frac{\pi}{n}}$$ $$ a_n = \frac{1}{n \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n}$$ \begin{align} \lim_{n\to\infty} \frac{a_n}{n} &= \lim_{n\to\infty} \frac{1}{n^2 \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\ &= \lim_{n\to\infty} \frac{1}{n \sin\frac{\pi}{n}} \cdot \lim_{n\to\infty} \frac{1}{n} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\ &= \frac{1}{\pi} \cdot \int_0^1 \sin(\pi x) dx = \\ &= \frac{1}{\pi} \cdot \frac{-1}{\pi}\cos(\pi x)\big|_{x=0}^{x=1} =\\ &= \frac{1}{\pi}\cdot\frac{2}{\pi} = \frac{2}{\pi^2} \end{align}