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 n&gt; K(e)[/tex] then |x_n_L|&amp;lt; \epsilonyou mean &quot;for all n&gt; N(e).<br /> That only has to be true for <b>some</b> n&gt; 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...