Find a value of $\;\lim\limits_{n\rightarrow\infty}\frac{a_{n}}{n}$ 
(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.
 A: 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}
A: Since you ask for a shortcut, here is what I have got--
Since we are talking about an infinite polygon, we may very well express it as a circle under limiting conditions. (A nice animation Here )
Hence, the average value of the length of diagonals becomes the average value of the length of line segments drawn from the circle to all of its points. 
There are an infinite number of these lines, and hence we need to find an expression for their lengths. For that, we assume that in a cartesian plane, a circle of radius r is centred at the origin, with one of our endpoints along the y axis. 
Consider any point P. the polar coordinates of P are $(rsin\theta,rcos\theta)$ while that of A are $(-r,0)$
AP=$\sqrt{(r+r cos\theta)^2+(r sin\theta)^2}=r\sqrt{2(1+cos\theta)}$
For finding the values of average, as continuous values are present, we integrate with respect to from $(0,\pi)$ and divide--
average value=$r . \frac{\int _0^{\pi }\left(\ \sqrt{2\left(1+\cos \theta \right)}\right)d\theta }{\int _0^{\pi }d\theta }=r . \frac{4}{\pi}$
Now the circumference is n. That is, $n=2\pi r$ we substitute value of r and get the desired answer

$a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$

Hence proved................

Hope that my humble attempt provided you the 'shortcut' required....
If there are  more queries regarding this please do ask, and I would be happy to help :~)
