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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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    Answers (A) and (C) are the same.2012-07-17
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    Trigonometric functions will probably be tedious; think geometrically.2012-07-17
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    @joriki (C) will be 2 phi but i don't know how to write phi i.e 22/72012-07-17
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    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?2012-07-17
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    Ranabir: Can you please answer yes or no if a $\pi$ was supposed to be in the complex exponentials, like $e^{2\color{Red}\pi ik/n}$?2012-07-17
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    yes. the complex exponent $\pi$2012-07-17
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    Okay. Please verify my edit is correct: I put $\pi$ in the exponents. For future reference, if you don't know how to correctly state a problem, *say so* so that someone can fix it. The alternative is to invite answers to the wrong problem.2012-07-17
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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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    If you omitted $\pi$ in the exponents then this answer will be edited accordingly.2012-07-17
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    Ultimately what will be the correct choice of this problem and why please explain2012-07-17
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    @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 1.2012-07-17
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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