Multiple-choice question regarding $\lim\limits_{n \to \infty} \sum\limits_{k = 1}^n \left| e^{\frac{2\pi ik}{n}} − e^{\frac{2\pi i(k-1)}{n}} \right|$ 
The limit
  $$\lim_{n \to \infty} \sum_{k = 1}^n \left| e^{\frac{2\pi ik}{n}} − e^{\frac{2\pi i(k-1)}{n}} \right|$$                   is
(A) $2$ 
(B) $2e$ 
(C) $2\pi$ 
(D) $2i$.

I can't solve this problem. Do I need to use 
$$e^{i\theta} = \cos \theta + i \sin \theta$$
or do I need some other formula to proceed? I don't understand that is I need to interchange the limit and summation. Please help me. This is a multiple choice question from a sample test paper of ISI MSTAT examination.
 A: The segment from $e^{2\pi i(k-1)/n}$ to $e^{2\pi ik/n}$ is a segment along the interior of the unit circle. The collection from $k=1$ to $k=n$ spans the the whole circle from $0$ to $2\pi$ radians, so the sum of their lengths limits to the length of the circle of radius $1$.
The diagram below is for $n=15$. The red segments approximate the arc from $0$ to $2\pi$ radians; that is, $e^{0i}$ to $e^{2\pi i}$:
$\hspace{4.5cm}$
A: Route 1: 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 distances between consecutive vertices $|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}|$ are the side lengths and the sum is the perimeter of the $n$-gon, which will approximate the unit circle as $n\to\infty$. What is the circumference of the unit circle? Here's a visual aid with $n=5$ and $n=12$:

Route 2: We have
$$\sum_{k=1}^n|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}|=\sum_{k=1}^n|e^{2\pi ik/n}||1-e^{-2\pi i/n}| \\[5pt] =n|1-e^{-2\pi i/n}|.$$
The limit of this as $n\to\infty$ can be evaluated analytically by invoking a Taylor series expansion of the exponential function, $e^x\approx 1+x$ as $x\approx 0$ (formally, $e^x=1+x+O(x^2)$). Specifically,
$$\lim_{n\to\infty}n|1-(1-2\pi i/n+\cdots)|=? $$
A: $$\left|e^{\frac{i2k\pi}{n}}-e^{\frac{i2(k-1)\pi}{n}}\right|=\left|e^{\frac{i2k\pi}{n}}\right|\left|1-e^{\frac{i2\pi}{n}}\right|=\left|1-e^{\frac{i2\pi}{n}}\right|=\sqrt{(1-\cos\frac{2\pi}{n})^2+\sin^2\frac{2\pi}{n}}=2\sin\frac{\pi}{n}$$
$$\lim_{n\to\infty}\sum_{k=1}^n\left|e^{\frac{i2k\pi}{n}}-e^{\frac{i2(k-1)\pi}{n}}\right|=\lim_{n\to\infty}\sum_{k=1}^n2\sin\frac{\pi}{n}=2\lim_{n\to\infty}n\sin\frac{\pi}{n}=2\pi\lim_{n\to\infty}\frac{\sin\frac{\pi}{n}}{\frac{\pi}{n}}=2\pi$$
