Solving Series Convergence: ln(x)^ln ln(x)

[pierce15]

Homework Statement

Is this series convergent or divergent?

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}}

Homework Equations

ln(x)<x^a, \, a>0

The Attempt at a Solution

I don't see a way to tackle this one other than the comparison test.

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} < or > ?

What would I compare this to?

 

[Ray Vickson]

Homework Statement

Is this series convergent or divergent?

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}}

Homework Equations

ln(x)<x^a, \, a>0

The Attempt at a Solution

I don't see a way to tackle this one other than the comparison test.

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} < or > ?

What would I compare this to?

Re: LaTeX code: you need }} instead of } to close the \frac command properly. Also, you should write '\ln' instead of 'ln'; 'ln' produces $ln(n)$, while '\ln' gives $\ln(n)$.

 

[STEMucator]

Homework Statement

Is this series convergent or divergent?

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}}

Homework Equations

ln(x)<x^a, \, a>0

The Attempt at a Solution

I don't see a way to tackle this one other than the comparison test.

\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} < or > ?

What would I compare this to?

Slow down and take your hint into consideration.

For $a>0$, if $ln(n) < n^a$ then $(ln(n))^{ln(ln(n))} < (n^a)^{ln(ln(n))}$.

Reciprocating both sides, you get $\frac{1}{(ln(n))^{ln(ln(n))}} > \frac{1}{n^{aln(ln(n))}} >$ something I will allow you to notice for yourself.

Then you will have to break down the cases for positive a... you'll need two cases in total. Pretty sure it was 'p-series' though, not 'a-series' :).

 

[I like Serena]

Slow down and take your hint into consideration.

For $a>0$, if $ln(n) < n^a$ then $(ln(n))^{ln(ln(n))} < (n^a)^{ln(ln(n))}$.

Reciprocating both sides, you get $\frac{1}{(ln(n))^{ln(ln(n))}} > \frac{1}{n^{aln(ln(n))}} >$ something I will allow you to notice for yourself.

Then you will have to break down the cases for positive a... you'll need two cases in total. Pretty sure it was 'p-series' though, not 'a-series' :).

Based on visual inspection it looks as if $\frac{1}{(\ln n)^{\ln(\ln n)}} > \frac 1 {n^{0.1}} > \frac 1 n$ for n sufficiently large.
The series diverges.


I don't see how I can match that with your observation though.
The series $\dfrac{1}{n^{a\ln(\ln n)}}$ converges for any a>0, meaning it won't give conclusive evidence.