solving the limit $\lim_{n\to \infty}\sum_{k=1}^n|e^{(2πik)/n}-e^{(2πi(k-1))/n}$|

Question

$$\lim_{n\to \infty}\sum_{k=1}^n\left|e^{(2πik)/n}-e^{(2πi(k-1))/n}\right|$$

i can solve it geometrically. but is there any way to solve it using Euler's formula ?,

the answer will be one of these four :

a)$2\quad$ b)$2e\quad$ c)$2π\quad $ d)$2i$

Answer 1

We have \begin{align} \left\vert e^{2\pi ik/n} - e^{2\pi i(k-1)/n}\right \vert & = \left\vert e^{2\pi i(k-1)/n} \right \vert \left \vert e^{2\pi i/n} - 1\right \vert = \left \vert \cos\left(\dfrac{2\pi}n\right)-1 + i \sin\left(\dfrac{2\pi}n\right)\right \vert\\ & = \left \vert -2\sin^2\left(\dfrac{\pi}n\right) + 2i \sin\left(\dfrac{\pi}n\right) \cos\left(\dfrac{\pi}n\right) \right \vert = 2\sin\left(\dfrac{\pi}n\right) \end{align} Hence, your limit is $$\lim_{n \to \infty} \sum_{k=1}^n2\sin\left(\dfrac{\pi}n\right) = \lim_{n \to \infty} 2n \sin\left(\dfrac{\pi}n\right) = 2\pi$$

Answer 2

If $\gamma:[a,b]\to \mathbb {C}$ is a $C^1$ curve, then

$$\lim_{n\to \infty} \sum_{k=1}^{n}\left |\,\gamma \left (a+k\cdot \frac{b-a}{n}\right )-\gamma \left (a+(k-1)\cdot\frac{b-a}{n}\right )\right |$$

is the arc length of $\gamma.$ In the case of $\gamma :[0,2\pi] \to \mathbb {C}$ given by $\gamma (t) = e^{it},$ this is the arc length of the unit circle, namely $2\pi.$