Cauchy sequence

[triclino]

prove using the ε-definition,that : ln n is not a Cauchy sequence

[ajb]

Let me just define the notion of a Cauchy sequence for the benefit of our other users.

 

Definition

Let [math](X,d)[/math] be a metric space. A sequence of points [math]x_{1},x_{2}, \cdots[/math] is said to be a Cauchy sequence if and only if for every real number [math]\epsilon > 0[/math] there exists a positive integer [math]N[/math] such that for all (natural numbers) [math]n,m > N[/math] the distance function satisfies

 

[math]d(x_{n},x_{m})< \epsilon[/math].

 

In the opening question I think we can assume we are dealing with Euclidean space, thus [math]d(x_{n},x_{m}) = |x_{n}-x_{n}|[/math].

 

The intuitive idea is that terms in the sequence get closer and closer together. However, this in general does not mean that a limit exists in [math]X[/math].

 

So, triclino how far have you got with this problem?

[triclino]

Hence if ln n is not a Cauchy sequence:

 

 

there exists an ε>0 ,such that :

 

For all natural Nos N ,there exist natural Nos [math]n\geq N,m\geq N[/math] ,such that : [math] |ln n-ln m|\geq\epsilon[/math]

 

Now the question is how do we find ε,n,m to satisfy the above?

[uncool]

Hence if ln n is not a Cauchy sequence:

 

 

there exists an ε>0 ,such that :

 

For all natural Nos N ,there exist natural Nos [math]n\geq N,m\geq N[/math] ,such that : [math] |ln n-ln m|\geq\epsilon[/math]

 

Now the question is how do we find ε,n,m to satisfy the above?

 

Remember, it's not how to find epsilon, n, and m just like that. You have to find an epsilon such that for any N, you can find an n and m such that [stuff].

=Uncool-

[triclino]

Suggestion No 1:

 

For any natural No N

 

Let n = N and m = 2N,then |ln n-ln m| = |lnN- ln2N| = |lnN-ln2-lnN|= |ln2|

 

Hence ,choose : [math]0<\epsilon\leq |ln2|[/math]

 

Any other suggestions?

 

Apart from those where : m= 3N,4N,5N...............................kN?

[the tree]

Hence ,choose : [math]0<\epsilon\leq |ln2|[/math]

ε is kind of chosen for you, it should really be the first line of your proof. You're along the right lines you just need to put your thoughts in order. Edited January 18, 2010 by the tree