College Math Teaching

October 25, 2013

Why the sequence cos(n) diverges

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.).

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