# Negation of definition of convergence

1. Oct 26, 2013

### phospho

The definition of convergence is given by : $\forall \epsilon > 0, \exists N \in \mathbb{R}$ such that $|x_n - l | < \epsilon$ $\forall n \in \mathbb{N}$ with $n > N$

negate this statement and prove that the sequence $x_n = (-1)^nn$ is divergent using only the negation of the definition of convergence

ok here's my attempt of the negation of the statement:

$\exists \epsilon > 0 \forall N \in \mathbb{R}$ s.t. $\exists n \in \mathbb{N} n > N$ and $|x_n - l | \geq \epsilon$

proof that $x_n = (-1)^nn$ does not converge:

since the negation of the definition is "$\forall N$" let N = 1, and find $n_1 > 1$ then let $N = n_1$ and find $n_2 > N=n_1$ etc leaving us with a sequence of indicies $n_1 < n_2 < n_3..$

so lets prove that it does not converge to 1:

choose $epsilon = 1/2$ then for $N>0$ let n be the odd integers i.e. $n_1 = 1 < n_2 = 3 < n_3 = 5 ...$ then n > N hence $| (-1)^1(1) - 1 | >= 1/2$ which is true hence $x_n$ does not converge to 1

take $l > 1$ and choose $\epsilon$ s.t. $\epsilon < l - 1$ again, choosing $N > 0$ and n being all the odd integers greater than 0 we have n > N and $|(-1)(1) - l | \geq \epsilon$ which is true

take $l < 1$ and choose n to be all the odd integers above 0 then n > N and $|-1 - l | >= \epsilon$ which is true therefore $x_n$ does not converge

I think this works, but truth be told, I don't understand the negation of this statement. I mean, I understand HOW I got the negation, but I don't know what it means. The choosing N > 0 is from my lecture notes and I don't understand why that works, and why we can just choose N > 0 if the definition states for all N

2. Oct 26, 2013

### brmath

You are on the right track but I think it is a little simpler. Given the sequence $a_n = n(-1)^n$ you want to show it cannot have a limit. If it does have a limit L then for any $\epsilon$ there is an $N$ such that |$a_n -L$| whenever n > N.

Choosing $\epsilon$ = 1/2 is a great idea. Now what happens to $a_n$ as n gets large? How does each term relate to the previous one?

So you want to say something like: no matter what N I choose I can't get |$a_n -L$| < 1/2 for every n > N. Then explain why not.

The positive statement is I can find an N; the negation is I cannot find an N.