# Epsilon- N Proof a Sequence Diverges

1. Sep 1, 2015

### B3NR4Y

1. The problem statement, all variables and given/known data
1. Let $x_n = \frac{n^2 - n}{n}$ does $x_n$ converge or diverge?
2. Let $x_n = \frac{(-1)^n +1}{n}$ does $x_n$ converge or diverge

2. Relevant equations
A sequence converges if $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $n\geq N$ implies $| x_n - L | < \epsilon$
A sequence diverges if $\exists \epsilon_{0} > 0$, $\forall N \in \mathbb{N}$, $\exists n_{0} \geq N$ such that $|x_{n_{0}} - L | > \epsilon_{0}$

3. The attempt at a solution
For one, I noticed that $x_n = n-1$ which obviously diverges. But as my teacher says, I gotta prove it. So I take $\epsilon_0 = \frac{1}{2}, \, \, L \in \mathbb{R}, \, \, n \geq N, \, \, N \in \mathbb{N}$
$$1.) \, \, | (N - 1) - L | < \frac{1}{2}, \, \, \, \, \, \, \, \, |((N+1)-1) - L | = |N-L| <\frac{1}{2}$$

I can apply the triangle inequality to find that $|N-1-L| - |N-1| \le 1$, but I can see from 1.) that $|N-1-L| - |N-L| < 0$ Therefore $x_n$ diverges.

Obviously I wouldn't be posting here if that satisfied me, it doesn't. $|N-1-L| - |N-L|=(-\infty, \, 0)$ would satisfy both of those.

2.) I noticed that $x_n = \frac{(-1)^n +1}{n} > \frac{(-1)^n}{n}$ so all I should have to do is prove $b_n = \frac{(-1)^n}{n}$ diverges and that should be sufficient. So if we take $\epsilon_{0} = \frac{1}{2}, \, \, L = 1 \, \, n\geq N$, $|b_{n} - 1| > \epsilon$, $b_n = 1$ when n is even and $b_n = -1$ when n is odd, therefore $|-1 -1| = |-2| = 2 > \epsilon$ which proves $b_n$ is divergent therefore proving $x_n$ is divergent..

This doesn't satisfy my either, I understanding proving convergence just fine but for some reason I find proving divergence harder.

2. Sep 1, 2015

### Geofleur

Well, if a sequence diverges, then there will be some $\epsilon$ such that, no matter how big an $N$ you pick, $n \geq N$ will not make $|x_n - x | < \epsilon$. That is, we will have $|x_n - x| \geq \epsilon$ no matter how big $n$ gets. Does phrasing it that way help clarify things?

3. Sep 1, 2015

### B3NR4Y

Not particularly, I understand what the symbols mean, but actually executing a proof of divergence trips me up. I've started over on the first one, but choosing the epsilon and then proving that no matter what, $|x_n - L| > \epsilon$, confuses me.

4. Sep 1, 2015

### Geofleur

For your first series, you noticed that $x_n = n - 1$. So the first few terms of the series (assuming we start at $n = 1$) are 0, 1, 2, etc. Suppose $\epsilon$ is, say, 1/2. For the sequence to converge to x, we would need all the $x_n$ for big enough $n$ to make | x_n - x | < 1/2. We have the opposite situation - if you make $n$ big enough, | x_n - x | = | n - 1 - x | will get bigger than any finite number you choose!

5. Sep 1, 2015

### B3NR4Y

Oh! Okay, that makes a lot of sense.

For the second one, if we choose epsilon to equal 1

$$|\frac{(-1)^{n} +1}{n} - L| < 1$$
If n is odd, $\frac{L}{n} < 1$, which isn't necessarily true for all n. L has to be chosen so that $\frac{1}{L} > \frac{1}{n}$, but L can't be conditioned by n, it has to be fixed. So it's not looking good for $x_n$

If n is even, $\frac{|Ln-2|}{n} < 1$, which isn't true, since Ln - 2 ≥ n, so the fraction $\frac{|Ln-2|}{n}\geq 1$ Therefore $x_n$ diverges.

6. Sep 1, 2015

### Geofleur

Write down a bunch of terms of the sequence. Does it look like they are approaching a definite number?

7. Sep 1, 2015

### B3NR4Y

For the second one, no, it's either 0 or 1/n ∀n

But sadly just writing terms isn't sufficient for my professor.

8. Sep 1, 2015

### Geofleur

What happens to 2/n when n gets bigger and bigger?

9. Sep 1, 2015

### B3NR4Y

It also goes to zero, oops. So it converges and I have to prove that.

10. Sep 1, 2015

### Geofleur

You can't prove something that is false, after all :-) That's why it's good to actually write some concrete numbers and get a feel first.

11. Sep 1, 2015

### B3NR4Y

Thank you so much! I understand it better now, for sure. :)