# Proof of convergence

1. Oct 18, 2013

### converting1

Use only the definition of convergence to prove $\dfrac{cos(n)}{n} \rightarrow 0$

my proof:

given $\epsilon > 0$ $\exists N \in \mathbb{R}$ s.t. n>N => $|\dfrac{cos(n)}{n}| < \epsilon$

$|cos(n)| \leq 1$ therfore $\dfrac{|cos(n)|}{|n|} \leq \dfrac{1}{|n|} < \epsilon$ so $|n| > \dfrac{1}{\epsilon}$
so choose big N to be 1/e and we get $|\dfrac{cos(n)}{n}| < \epsilon$

is this proof ok?

from here: $\dfrac{|cos(n)|}{|n|} < \epsilon$ we can do $-\epsilon |n| < cos(n) < \epsilon |n|$ and if we consider $cos(n) < \epsilon |n|$ we get $\epsilon |n| > -1$ and therefore $n > \dfrac{-1}{\epsilon}$ and $n < \dfrac{1}{\epsilon}$ why do I get two different results contradicting with my first proof, i'm assuming i've gone wrong so if someone could correct me where, please.

2. Oct 18, 2013

### converting1

similarly if I was to prove it using the sandwich theorem, would it be ok in saying $-1 \leq cos(n) \leq 1$ and then dividing by n? i.e. $-1/n \leq cos(n)/n \leq 1/n$ seeing as we don't know if n is negative or not?

3. Oct 18, 2013

### LCKurtz

That is a statement of what you are to prove.

You have the idea but it would be better to rewrite it in a different order:

Suppose $\epsilon > 0$. Pick $N$ so that $N>\frac 1 \epsilon$. Then if $n > N$, then $n>\frac 1 \epsilon$ so$$\frac {|\cos(n)|} {n} \le \frac 1 n <\epsilon$$That's all you need to write.

4. Oct 18, 2013

### LCKurtz

You are going to have $\epsilon > 0$ and $N > \frac 1 \epsilon$, so your $n$ will be positive if $n>N$.

5. Oct 18, 2013

### brmath

You didn't say, but from the way you did the proof can we assume n $\rightarrow \infty$? If so, the proof is fine, altho it seems to have a typo where you wrote "e" instead of $\epsilon$.
It's |cos(n)|/|n| < $\epsilon$. If n$\rightarrow \infty$ then n > 0, and you don't need the absolute values on n. So you have |cos(n)| < n$\epsilon$ where n >1/$\epsilon$. This means only that $\epsilon$n > 1. So all we are saying here is that |cos(n)| is less than a number which is > 1. That's perfectly fine. Where we would be in trouble is if we concluded that n$\epsilon$ < 1, because then it wouldn't be true that |cos(n)| < n$\epsilon$ for every n.

The computation you did somehow got the inequalities mixed up. You should have -$\epsilon$n < -1, because you are comparing with the smallest value of cos(n).

Questioning things this way is very good. It forces you to get things absolutely clear, rather than half understanding.

6. Oct 18, 2013

### converting1

continuing from $-\epsilon|n| < cos(n) < \epsilon|n|$ and looking at the upper inequality only i.e. $cos(n) < \epsilon|n|$ and you said we take the smallest value of cos(n) which is -1 right? so $\epsilon |n| > -1$ so $n > -1/\epsilon$ ? If, however, we take the largest value of cos(n) we get the correct answer. Which ones should we be taking if we had $-\epsilon|n| < cos(n) < \epsilon|n|$

7. Oct 18, 2013

### converting1

yes but assuming I did not prove that $n > 1/\epsilon$ then how would I know it's ok to divide by n?

8. Oct 18, 2013

### brmath

When I said "smallest value of the cosine" I was looking at the left hand inequality. Trying to explain further:

The cosine as you know oscillates between 1 and -1. -n$\epsilon$ < cos(n) for every n requires that -n $\epsilon \le -1$. You have chosen your n > N so that inequality is true. Similarly cos(n) < n$\epsilon$ for every n would require n$\epsilon$ > 1. Again, because of your choice of n that is true.

Once you've chosen an n so that |cos(n)|/n < $\epsilon$, all other arithmetic restatements of it have to be true. The restatement you have given here is certainly true because of your choice of n vs the $\epsilon$ that was chosen. Your statement is not useful for convergence purposes, but does serve to force us to think about what these inequalities really mean, and that is a good thing.

Do you understand it better now? Or do we need more discussion?

9. Oct 18, 2013

### LCKurtz

I'm not sure why this is an issue for you. You are proving a limit as $n\to\infty$. You are really only interested in large positive $n$.

10. Oct 19, 2013

### brmath

As $\epsilon$ gets smaller and smaller, N gets larger and larger. If you want to prove the series converges you have to choose an N, and you properly chose it so that N > 1/$\epsilon$, which means for n>N n >1/$\epsilon$. So there is certainly no worry about dividing by n.

You didn't "prove" n >1/$\epsilon$. You chose it in order to show that |cos(n)|/n goes to 0 as n $\rightarrow \infty$. So no matter what $\epsilon$ you choose there is always an N so that for every n > N |cos(n)|/n < $\epsilon$.

I think it might help you to review again the definition of convergence as n $\rightarrow \infty$. I get the impression you haven't quite grasped the idea, even though you did the computation correctly.