Relating a geometric limit deriving the arc length of a polar curve to a Riemann sum So recently one of my friends came to me with this problem
You start with a circle of radius 1 which is divided into n subdivisions. As you go counterclockwise, you plot a point which is +(1/n) radially outward compared to the last point. So for the case n=8, you start at r=0, and then at the next subdivision, you plot the point 1/8. At the next subdivision you plot 2/8, then 3/8, and so on until you reach 1. He wanted to find the length of the curve formed as n approaches infinity
He started by looking at the triangle formed by the connected points. Let $C_{n}$ be the length of the segment formed by the connected points inside a sector. Then $\frac{k}{n}$ is the length of the segment from the center of the circle to the start point of $C_{n}$. In addition $\frac{k+1}{n}$ is the length of the segment from the center of the circle to the endpoint of $C_{n}$. Using law of cosines, $C_{n}$ can be expressed as this:
$$C_{n}=\sqrt{\left(\frac{k}{n}\right)^{2}+\left(\frac{k+1}{n}\right)^{2}-2\left(\frac{k}{n}\right)\left(\frac{k+1}{n}\right)\cos\left(\frac{2\pi}{n}\right)}$$
So in a discrete case, the length of the curve for n subdivisions is
$$\sum_{k=1}^{n}\sqrt{\left(\frac{k}{n}\right)^{2}+\left(\frac{k+1}{n}\right)^{2}-2\left(\frac{k}{n}\right)\left(\frac{k+1}{n}\right)\cos\left(\frac{2\pi}{n}\right)}$$
Taking the limit as n approaches infinity, and by doing some factoring, the length of the continuous curve for an infinite amount of subdivisions can be written as
$$\lim_{n \to {\infty}} \frac{1}{n}\sum_{k=1}^{n}\sqrt{k^{2}+\left(k+1\right)^{2}-2k\left(k+1\right)\cos\left(\frac{2\pi}{n}\right)}$$
Which is approximately 3.38
When I first saw this sum, It made me think about Riemann sums. The idea of taking a sum up to n, and them dividing it by n and taking the limit to infinity feels very closely connected to the definition of an integral. I started looking for a way to connect this infinite sum to an integral somehow.
Finding the closed form:
Intuitively it could be figured out that the equation of the continuous curve is $r=\frac{1}{2\pi}\theta$. Using the polar curve length formula, it can be shown that the length of the continuous curve is:
$$\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{1+x^{2}}dx$$
This can be expressed as an infinite sum using the definition of an integral using a Riemann sum:
$$\lim \limits_{n \to +\infty} \frac{b-a}{n} \sum_{k=1}^{n} f\Big( a + k\frac{b-a}{n} \Big) = \int_{a}^{b} f(t) \; dt.$$
Thus the integral is equal to:
$$\lim_{n \to {\infty}} \frac{1}{n}\sum_{k=1}^{n}\sqrt{\left(\frac{2\pi k}{n}\right)^{2}+1}$$
The 2pi cancels out in the end, which makes this infinite sum look very similar to the one we derrived earlier. With a little algebra, the infinite sum derrived from the discrete case can be rewritten as:
$$\lim_{n \to {\infty}} \frac{1}{n}\sum_{k=1}^{n}\sqrt{\left(2k^{2}+2k\right)\left(1-\cos\left(\frac{2\pi}{n}\right)\right)+1}$$
They both look very similar and they both converge experimentally to 3.38 :)
I'm tempted to assume from this equation that $\sqrt{\left(2k^{2}+2k\right)\left(1-\cos\left(\frac{2\pi}{k}\right)\right)} = \frac{2\pi k}{n}$, but I don't think I'm allowed to do that :/. There has to be a rule stopping me from doing that. Experimentally they don't really look equal, so I hit a wall with this method.
So I guess my final question is whether there is some way to rewrite that initial infinite sum as a more easily readable Riemann sum. After that, is there some way to write that as a neat integral? Thanks!
 A: Using Taylor expansions, for small $x$:
$$\cos x=1-\frac {x^2}2  +\mathcal O(x^4) \tag{1}$$ and
$$\sqrt{1+x+\mathcal O(x^2)}=\sqrt{1+x}+\mathcal O(x^2)\tag{2}$$
With that:
$$\begin{split}
\frac{1}{n}\sum_{k=1}^{n}\sqrt{\left(2k^{2}+2k\right)\left(1-\cos\left(\frac{2\pi}{n}\right)\right)+1} &= \frac 1 n \sum_{k=1}^n \sqrt{\left(2k^{2}+2k\right)\left(\frac {2\pi^2}{n^2} +\mathcal O\left(\frac 1 {n^4}\right)\right)+1}\\
&=  \frac 1 n \sum_{k=1}^n \sqrt{1 + \frac{4\pi^2k^2}{n^2} +\mathcal O\left(\frac 1 {n}\right)}\\
&=  \frac 1 n \sum_{k=1}^n\left (\sqrt{1 + \frac{4\pi^2k^2}{n^2}} +\mathcal O\left(\frac 1 {n}\right)\right)\\
&=  \left(\frac 1 n \sum_{k=1}^n\sqrt{1 + \frac{4\pi^2k^2}{n^2}}\right) +\mathcal O\left(\frac 1 {n}\right)\\
&\longrightarrow \int_0^1\sqrt{1+4\pi^2x^2}dx \simeq 3.38...
\end{split}$$
A: Staring from @Stefan Lafon's elegant solution, we have
$$\int_0^1\sqrt{1+4\pi^2x^2}\,dx=\frac{1}{2} \sqrt{1+4 \pi ^2}+\frac{\sinh ^{-1}(2 \pi )}{4 \pi }\approx 3.3830443$$
Now, just computing
$$S_n=\frac{1}{n}\sum_{k=1}^{n}\sqrt{\left(2k^{2}+2k\right)\left(1-\cos\left(\frac{2\pi}{n}\right)\right)+1}$$ for $n=10^p$, we have
$$\left(
\begin{array}{cc}
p & S_{10^p} \\
 1 & 3.8679722 \\
 2 & 3.4363000 \\
 3 & 3.3884020 \\
 4 & 3.3835805 \\
 5 & 3.3830980 \\
 6 & 3.3830497 \\
 7 & 3.3830449 \\
 8 & 3.3830444
\end{array}
\right)$$
