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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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    @Ranabir : See the difference between [previous question ]http://math.stackexchange.com/questions/172305/find-lim-n-rightarrow-infty-sqrtna-n1-a-n-where-a-n-frac1) and this one. I noticed a drastic improvement. Good.2012-07-19

2 Answers 2

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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 $n=12$:

ngon

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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    @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.2012-07-17
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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 $e^{2\pi i}$:

$\hspace{4.5cm}$enter image description here