1. ## Infinite Series (13).

Determine the Convergence/Divergence of the following series:
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$

Sure, It will be solved by the comparison tests.

But its hard to find a series which works in in these tests.

And sorry for my bad english

2. Originally Posted by TWiX
Determine the Convergence/Divergence of the following series:
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln(n)}}$

Sure, It will be solved by the comparison tests.

But its hard to find a series which works in in these tests.

And sorry for my bad english
I know you will hate this...but Cauchy's condensation test. $\displaystyle \frac{2^n}{\ln^{\ln(2^n)}\left(2^n\right)}=..$

Alternatively, $\displaystyle \frac{1}{\ln^{\ln(n)}(n)}=\text{exp}\left(\ln\left (\frac{1}{\ln^{\ln(n)}(n)}\right)\right)=\text{exp }\left(-\ln(n)\ln(\ln(n))\right)$$\displaystyle =\text{exp}\left(\ln\left(\frac{1}{n^{\ln(\ln(n))} }\right)\right)=\frac{1}{n^{\ln(\ln(n))}}, and clearly \displaystyle n^{\ln(\ln(n))}\geqslant n^2\text{, }n\geqslant e^{e^2}. 3. Originally Posted by Drexel28 I know you will hate this...but Cauchy's condensation test. \displaystyle \frac{2^n}{\ln^{\ln(2^n)}\left(2^n\right)}=.. Alternatively, \displaystyle \frac{1}{\ln^{\ln(n)}(n)}=\text{exp}\left(\ln\left (\frac{1}{\ln^{\ln(n)}(n)}\right)\right)=\text{exp }\left(-\ln(n)\ln(\ln(n))\right)$$\displaystyle =\text{exp}\left(\ln\left(\frac{1}{n^{\ln(\ln(n))} }\right)\right)=\frac{1}{n^{\ln(\ln(n))}}$, and clearly $\displaystyle n^{\ln(\ln(n))}\geqslant n^2\text{, }n\geqslant e^{e^2}$.
Yes i hate that cauchy
nerd man .. replace every n by 2^n and check for the new one
OMG
it makes it two problems not only one
NERD CAUCHY!

Thanks for the second solution.
I got it
am math professor
and omg, you did not get a headache when you writing this in latex ?
Many exponents
Thank you ^_^