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.

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Answers (A) and (C) are the same. – joriki Jul 17 '12 at 15:55
Trigonometric functions will probably be tedious; think geometrically. – anon Jul 17 '12 at 15:59
@joriki (C) will be 2 phi but i don't know how to write phi i.e 22/7 – Argha Jul 17 '12 at 15:59
I think you mean $\pi$ (which is approximately $22/7$, but not exactly). The $\TeX$ command for $\pi$ is \pi (not \phi, which produces the Greek letter phi, $\phi$). To use $\TeX$, enclose the code in single dollar signs for inline formulas, or in double dollar signs for displayed equations. By the way, it would have made sense to point out the fact that you omitted part of the problem because you couldn't write it. Also, I wonder, if you omitted it there, did you perhaps also omit it in the exponents? – joriki Jul 17 '12 at 16:03
Ranabir: Can you please answer yes or no if a $\pi$ was supposed to be in the complex exponentials, like $e^{2\color{Red}\pi ik/n}$? – anon Jul 17 '12 at 16:10

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,

$$\lim_{n\to\infty}n|1-(1-2\pi i/n+\cdots)|=?$$

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If you omitted $\pi$ in the exponents then this answer will be edited accordingly. – anon Jul 17 '12 at 16:07
Ultimately what will be the correct choice of this problem and why please explain – Argha Jul 17 '12 at 16:27
@Ranabir: I have given you an explanation of how to find the answer. If there are parts you do not understand, speak up. For example, do you know what a Taylor expansion is? Do you know about $n$th roots of unity and how they form the vertices of a regular $n$gon in the unit circle? If you want to know what the correct choice is, see the rhetorical question at the end of Route 1. – anon Jul 17 '12 at 16:38
@Ranabir: If this is, as you say, merely an example question from a sample test, then what on earth do you think you will gain from simply being told the correct answer? Not being able to answer the question yourself means that you're missing some general knowledge that the test will expect you to have, and your goal should be to acquire that general knowledge, not the particular answer to this (in itself fairly irrelevant) question. – Henning Makholm Jul 17 '12 at 18:18

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 $e^{2\pi i}$:

$\hspace{4.5cm}$

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