Can You Interpret This Limit Geometrically?

Relevant wiki: Roots of Unity

Geometrically, the $n^{th}$ roots of unity $e^{2\pi i k/n}$ form a regular $n$ -gon in the complex plane $\Bbb C$ . So the expression $|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}|$ just gives us the measure of the side of the $n$ -gon, so the obvious conclusion that can be drawn is that as $n\to\infty$ the $n$ -gon approaches the shape of a unit circle. To give a feeling of the $n$ -gon transforming into a circle as $n\to\infty$ , consider the following diagrams which make it clear that as $n\to\infty$ the $n$ -gon approaches to become a circle.

$n=5$

$n=10$

$n=13$

So, as $n\to\infty$ the polygon turns in to a circle and as the question asks for it perimeter, so from this point on it turns really easy and the solution comes out to be $2\pi$ the perimeter of the circle.

Alternative solution:-

We have

$\displaystyle \sum_{k=1}^n|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}|=\sum_{k=1}^n|e^{2\pi ik/n}||1-e^{-2\pi i/n}| \\[5pt] =n|1-e^{-2\pi i/n}|.$

From Taylor Series Expansion, we solve the limit as follows:-

$n|1-e^{-2\pi i/n}|=n \left| 1-1+\dfrac{2i\pi}{n} + O\left( \left(\dfrac{2i\pi}{n} \right)^2 \right) \right|=n \left| \dfrac{2i\pi}{n} \right| =2\pi$