# Definition of the limit of a sequence

• B
Homework Helper
2019 Award
Read it out to me, then!

"There exists ..."
The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and we can say for all N, that: n is larger than N and |un - L| is larger or equal to epsilon.

fresh_42
Mentor
The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and when we can say for all N, that: n > N and |un - L| is smaller or equal to epsilon.
You are repeating what you already said. There is a quantifier missing. It might help to negate what I've written in #24.

Homework Helper
2019 Award
You are repeating what you already said. There is a quantifier missing. It might help to negate what I've written in #24.
∃ε > 0, ∀N, ∃n>N: |un - L | >= ε

I believe this is the negation.

PeroK
Homework Helper
Gold Member
The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and when we can say for all N, that: n > N and |un - L| is smaller or equal to epsilon.
I still don't know what that means. As an antidote to this symbolic stuff:

A sequence does not converge (to L) if there is a finite distance from L such that no matter how far you go along the sequence, at some later point it will be further than this distance from L.

I would try to think about that geometrically as well: L as a line, $\epsilon$ as the width of a tunnel either side of L and the sequence (at least at some points) continually going outside the tunnel.

Whereas, if it converges to L, then no matter how narrow you make the tunnel, the sequence eventually goes into the tunnel and stays there!

fresh_42
Mentor
∃ε > 0, ∀N, ∃n>N: |un - L | >= ε

I believe this is the negation.
You should (in order to practice) apply this to Perok's sequence and show why it does not converge.

Svein
Given a sequence xn, we say that x is a cluster point for xn if $(\forall \epsilon >0)(\forall N)(\exists n>N)(\vert x-x_{n} \vert <\epsilon)$.
Given a sequence xn, we say that xn converges to x if $(\forall \epsilon >0)(\exists N)(\forall n>N)(\vert x-x_{n} \vert <\epsilon)$.