Proving Negation of Limit Definition

[antiemptyv]

Homework Statement

I'm trying to show that a sequence does not have a limit, so that would mean proving the negation of the limit definition is true, right? Is this a correct negation of the definition of what it means for a sequence to have a limit?

Homework Equations

The definition of the limit of a sequence $$(x_n)$$.
The sequence $$(x_n)$$ converges to $$L$$ if given $$\epsilon > 0$$, $$\exists K(e) \in \mathbb{N} \ni$$ if $$n > K(e)$$, then $$|x_n-L| < \epsilon$$.

The Attempt at a Solution

The limit of a sequence $$(x_n)$$ is not L if $$\exists \epsilon > 0 \ni \forall K \in \mathbb{N}$$, $$\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon$$.

 

[EnumaElish]

I think that is right, except it seems as if you have used too many N's, \in's or \ni's.

 

[HallsofIvy]

Homework Statement

I'm trying to show that a sequence does not have a limit, so that would mean proving the negation of the limit definition is true, right? Is this a correct negation of the definition of what it means for a sequence to have a limit?

Homework Equations

The definition of the limit of a sequence $$(x_n)$$.
The sequence $$(x_n)$$ converges to $$L$$ if given $$\epsilon > 0$$, $$\exists K(e) \in \mathbb{N} \ni$$ if $$n > K(e)$$, then $$|x_n-L| < \epsilon$$.

The Attempt at a Solution

The limit of a sequence $$(x_n)$$ is not L if $$\exists \epsilon > 0 \ni \forall K \in \mathbb{N}$$, $$\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon$$.

Not if by "if [itex]n> K(e)[/tex] then [tex]|x_n_L|< \epsilon[/itex] you mean "for all n> N(e).
That only has to be true for some n> Ke)

 

[antiemptyv]

Yes, it all seems right now I guess. Thanks! and oh yeah, i guess while editting, i left in a few extra symbols...