Determining the convergence/divergence of this series

[TogoPogo]

I am required to determine the convergence/divergence of the following series:
$$ \sum_{n = 2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$$

Which test should I use? Wolfram Alpha says that the comparison test was used to determine that it was convergent, but I have no idea what series I should compare it to.

It also said that the root/ratio tests gave inconclusive results.

Thanks!

 

[Mark44]

I am required to determine the convergence/divergence of the following series:
$$\sum\frac{1}{(lnx)^{lnx}}$$

from n=2 to n=infinity.

How does n play a role here?

It's a bit hard to see... but the series is 1/(lnx)^(lnx).
Mod note: Should be fixed now[/color]

Which test should I use? Wolfram Alpha says that the comparison test was used to determine that it was convergent, but I have no idea what series I should compare it to.

It also said that the root/ratio tests gave inconclusive results.

Thanks!

 

[TogoPogo]

Apologies, it should be as x approaches infinity...

Thanks for the fix!

 

[Mark44]

Then this is what you want.
$$ \sum_{n = 2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$$

 

[TogoPogo]

Then this is what you want.
$$ \sum_{n = 2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$$

That's correct

 

[pierce15]

Integral test! (joking)

First, rewrite the sum using $e^{\ln z} = z$. Then use the fact that after some finite value of $n$, $\ln n > a, \; a>0$. Here, I would just pick a value of $a$, let's use 2. Then:

$$\sum_{n=2}^{\infty} \frac{1}{{\ln n}^{\ln n} } = \sum_{n=0}^{\infty} \frac{1}{e^{\ln({\ln n}^{\ln n}) } } = \sum_{n=2}^{\infty} \frac{1}{e^{{\ln n}^{\ln \ln n}} } = \sum_{n=2}^\infty \frac{1}{n^{\ln \ln n}}$$

By some algebraic manipulations, $\ln n > 2, \, \forall n>e^2 \Longrightarrow \ln \ln n > \ln 2 \Longrightarrow n^{\ln \ln n} > n ^ {\ln 2} \Longrightarrow \frac{1}{n^{\ln \ln n}} < \frac{1}{n^{\ln 2}}$

Can you finish it from here?

 

[TogoPogo]

Integral test! (joking)

First, rewrite the sum using $e^{\ln z} = z$. Then use the fact that after some finite value of $n$, $\ln n > a, \; a>0$. Here, I would just pick a value of $a$, let's use 2. Then:

$$\sum_{n=2}^{\infty} \frac{1}{{\ln n}^{\ln n} } = \sum_{n=0}^{\infty} \frac{1}{e^{\ln({\ln n}^{\ln n}) } } = \sum_{n=2}^{\infty} \frac{1}{e^{{\ln n}^{\ln \ln n}} } = \sum_{n=2}^\infty \frac{1}{n^{\ln \ln n}}$$

By some algebraic manipulations, $\ln n > 2, \, \forall n>e^2 \Longrightarrow \ln \ln n > \ln 2 \Longrightarrow n^{\ln \ln n} > n ^ {\ln 2} \Longrightarrow \frac{1}{n^{\ln \ln n}} < \frac{1}{n^{\ln 2}}$

Can you finish it from here?

Thanks for your help. If I'm not mistaken, isn't $\frac{1}{n^{\ln 2}}$ divergent? Or is there another series that I should be comparing to?

 

[Dick]

Thanks for your help. If I'm not mistaken, isn't $\frac{1}{n^{\ln 2}}$ divergent? Or is there another series that I should be comparing to?

Yes, it is divergent. piercebeatz was maybe rushing it a little. Try to use the sort of things piercebeatz was using and try to work it out for yourself. Try comparing with a p series 1/n^p. And remember the comparison doesn't have to work for all n. Just for sufficiently large n. p=ln(2) is too small for convergence. There are other choices.

 

[pierce15]

Sorry, didn't realize that 2 was a bad choice for a. just let a=e^2 or something like that.

 

[Dick]

Sorry, didn't realize that 2 was a bad choice for a. just let a=e^2 or something like that.

I would give TogoPogo some time to work this out solo. But sure, your solution was fine in general.