Why the sequence cos(n) diverges

October 25, 2013

Why the sequence cos(n) diverges

Filed under: analysis, calculus, editorial, elementary number theory, number theory, point set topology, sequences — Tags: hard calculus problems — collegemathteaching @ 12:59 am

We are in the sequences section of our Freshman calculus class. One of the homework problems was to find whether the sequence $a_n = cos(\frac{n}{2})$ converged or diverged. This sequence diverges, but it isn’t easy for a freshman to see.

I’ll discuss this problem and how one might go about explaining it to a motivated student. To make things a bit simpler, I’ll discuss the sequence $a_n = cos(n)$ instead.

Of course $cos(x)$ is periodic with a fundamental region $[0, 2\pi]$ so we will work with that region. Now we notice the following:

$n (mod 2 \pi)$ is a group with the usual operation of addition.

By $n (mod 2 \pi)$ , I mean the set $n + k*2\pi$ where $k \in \{..-2, -1, 0, 1, 2, 3,...\}$ ; one can think of the analogue of modular arithmetic, or one might see the elements of the group $\{ r| r \in [0, 2 \pi), r = n - k 2\pi \}$ .

Of course, to get additive inverses, we need to include the negative integers, but ultimately that won’t matter. Example: $1, 2, 3, 4, 5, 6$ are just equal to themselves $mod 2 \pi.$ $7 = 7 - 2\pi (mod 2\pi), 13 = 13 - 4 \pi (mod 2\pi)$ , etc. So, I’ll denote the representative of $n (mod 2\pi)$ by $[n]$ .

Now if $n \ne m$ then $[n] \ne [m]$ ; for if $[n]=[m]$ then there would be integers $j, k$ so that $n + j2\pi = m +k2\pi$ which would imply that $|m-n|$ is a multiple of $\pi$ . Therefore there are an infinite number of $[n]$ in $[0, 2\pi]$ which means that the set $\{[n]\}$ has a limit point in the compact set $[0, 2\pi]$ which means that given any positive integer $m$ there is some interval of width $\frac{2\pi}{m}$ that contains two distinct $[i], [j]$ (say, $j$ greater than $i$ .)

This means that $[j-i] \in (0, \frac{2\pi}{m})$ so there is some integers $k_2, k_3,$ so that $k_2[j-i] \in (\frac{2\pi}{m}, \frac{2*2\pi}{m}), k_3[j-i] \in (\frac{2*2\pi}{m}, \frac{3*2\pi}{m})$ , etc. Therefore there is some multiple of $[j-i]$ in every interval of width $\frac{2\pi}{m}$ . But $m$ was an arbitrary positive integer; this means that the $[n]$ are dense in $[0,2\pi]$ . It follows that $cos([n]) = cos(n)$ is dense in $[-1,1]$ and hence $a_n = cos(n)$ cannot converge as a sequence.

Frankly, I think that this is a bit tough for most Freshman calculus classes (outside of, say those at MIT, Harvard, Cal Tech, etc.).

Comments (5)

5 Comments »

[…] a question about the sequence . That sequence diverges, but nailing down a proof is not trivial. Here is a discussion about ; the same principle applies […]

Pingback by 22 October lesson: introduction to sequences – MTH 122: Calculus II — October 22, 2020 @ 5:47 pm

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Oscillating sequences -1 to1

Comment by BU180212 SANTHAKUMAR V — August 21, 2021 @ 4:58 am

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- Your proof is nice and easy to follow, but u show much more than stated. To prove divergence only, using addition-theorems for cos(n+1), sin(2n) is much shorter and may be appropriate to freshman students.
  
  Comment by Tom — April 16, 2022 @ 4:35 am
  
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  - LOL..thanks. Sometimes when I see a way, I go directly there and don’t think of easier ways.
    
    Comment by oldgote — April 16, 2022 @ 11:06 am
Great! thank you very much

Comment by Ilay — August 9, 2022 @ 8:47 pm

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College Math Teaching

October 25, 2013