divergence to infinity

[triclino]

prove that the sequence ln n diverges to infinity

[uncool]

prove that the sequence ln n diverges to infinity

Define "diverging to infinity". Then try using that definition.

=Uncool-

[triclino]

Define "diverging to infinity". Then try using that definition.

=Uncool-

 

 

So given ε>0 ,what do you think should be the natural No N chosen ,so that :

 

for all natural Nos [math] n\geq N[/math] , ln n>ε

[insane_alien]

not sure what that second bit has to do with the first bit.

[the tree]

You only really need to pick a big-N if you're proving that something converges. For this I'd recomend finding a subsequence that is known to diverge and assert the lemma that a sequence with a divergent subsequence must be divergent itself.

[triclino]

not sure what that second bit has to do with the first bit.

 

According to the definition of divergence to infinity:

 

Given an ε>0 we have to find a natural No N ,such that:

 

for all [math]n\geq N\Longrightarrow ln n>\epsilon[/math].

 

So the central issue here is finding that N

[the tree]

Well no, it isn't. Look back at your definitions of convergence and divergence.

[triclino]

Well no, it isn't. Look back at your definitions of convergence and divergence.

 

Which definitions??

[the tree]

A sequence [imath]\{ x_n \}[/imath] is said to converge to a limit [imath]l[/imath] iff for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | < \epsilon[/imath].

 

Now, this is very important so read carefully: A sequence diverges if and only if it does not converge.

 

Got that? So you need to look at the negate of the definition of convergence.

 

To be really, horribly slow about this, the negate of:

 

"for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | < \epsilon[/imath]."

 

is not and never will be:

 

"for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | > \epsilon[/imath]."

[triclino]

I suggest you read the opening post again carefully .

 

 

It does not ask that the sequence ln n does not converge,

 

But,

 

 

That the sequence ln n diverges to infinity

 

Who said that the negation of convergence to a limit l is :

 

" for a given ε>0,[math]\exists N=N(\epsilon)[/math] such that [math]\forall n>N |x_{n}-l|>\epsilon[/math]"

 

Read all the posts again, nowhere you will find that.

 

A sequence [math] x_{n}[/math] diverges to infinity iff

 

for all ,ε>0 there exists a natural No N ,such that:

 

for all ,n: [math] n\geq N\Longrightarrow x_{n}>\epsilon[/math]

 

This is the definition given by K.G BINMORE in his book :Mathematical A nalysis ,on pages 38-39

 

To mention one of the books producing that definition