How do I prove that the sequence cos(2n)/2n is cauchy? I know I must pick an

Question

How do I prove that the sequence $\frac{\cos2n}{2n}$is cauchy (using the definition of Cauchy Sequences)? I know I must pick an $\epsilon$, but I don't know which one specifically to pick, or how to prove it. Any help would be greatly appreciated.

Answer 1

Informally, to prove that the sequence is Cauchy you must show that if you go really far in the sequence, the terms are getting closer and closer together.

Now for your sequence. For any $x$ we have $-1 \le \cos x \le 1$, so $-\dfrac 1{2n} \le \dfrac {\cos2n}{2n} \le \dfrac {1}{2n}$

the black dots are $\dfrac {\cos2n}{2n}$ and curves are $\pm \dfrac {1}{2x}$. Indeed as you go farther and farrther along in the sequence the points have to be between $-\dfrac {1}{2n}$ and $\dfrac {1}{2n}$, so they will be getting closer and closer to each other. Can you figure out how to make a rigorous argument out of this?

EDIT: (Please see the new image). Fix an $\epsilon > 0$. We want to find the point $P$ on the $x$-axis such that the vertical distance between $\pm \dfrac {1}{2x}$ is equal to (or less than) $\epsilon$.

If we find this point $P$, then we will know that anywhere past $P$ on the $x-$axis, the distance between any two points of our sequence will be less than $\epsilon$ (in the image, look at the segments of lengths $a$ and $b$. If we name the points they correspond to as $a_m$ and $a_n$, then $|a_m-a_n|$ is $a+b$), which will be less than $\epsilon$.

We know that this argument is gonna work because we can see it; so if we fiddle enough with the algebra, the proof should pop out sooner or later. The best way to do it is to try to mirror the geometric steps.

The first thing we need to do is to find that $P$. At any point $x_0$ on the $x-$ axis, the distance between the two curves is $\dfrac {1}{2x_0} - \left(-\dfrac {1}{2x_0} \right) = \dfrac {1}{x_0}$. So if at $P$ this distance is $\epsilon$, then $\dfrac {1}{P} = \epsilon \implies P = \dfrac {1}{\epsilon}$. Now of course we don't include this distance argument in the formal proof; in the proof we would just say "Fix an $\epsilon >0$ and let $N$ be some integer greater than $\dfrac {1}{\epsilon}$", and work only algebraically.

Now what is left to show is that if you take any $n, m$ to the right of $N$, then $|a_n-a_m| < \epsilon$. How would you convince someone of this geometrically? You would say that $a_m, a_n$ are between the curves, and the distance between the curves after $N$ is less than $\epsilon$, so the distance between $a_m, a_n$ is less than $\epsilon$. Now try to put it into algebra; let me know if you need any help.