Author 
Topic: Convergent or Divergent? (Read 2161 times) 

ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Convergent or Divergent?
« on: Nov 26^{th}, 2008, 8:31am » 
Quote Modify

(nlogn.loglogn)^{1} n=3

« Last Edit: Nov 26^{th}, 2008, 1:15pm by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Obob
Senior Riddler
Gender:
Posts: 489


Re: Convergent or Divergent?
« Reply #1 on: Dec 1^{st}, 2008, 9:29pm » 
Quote Modify

Integral test.


IP Logged 



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: Convergent or Divergent?
« Reply #2 on: Dec 4^{th}, 2008, 3:49pm » 
Quote Modify

on Dec 1^{st}, 2008, 9:29pm, Obob wrote: Ah yes, I hadn't realized that the integral of the given function is logloglogx. Perhaps I should have asked how many terms are required for the sum to exceed 10.


IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: Convergent or Divergent?
« Reply #3 on: Dec 15^{th}, 2008, 7:21pm » 
Quote Modify

on Dec 4^{th}, 2008, 3:49pm, ThudanBlunder wrote: Perhaps I should have asked how many terms are required for the sum to exceed 10. 
 Let N = Number of terms required for function to exceed 10 F(n): 1/loglog(n) for n > 2 N: 1 F(n): 1/log(n) for n > 1 N: 20 F(n): 1/n for n > 0 N: 33 F(n): 1/n for n > 0 N: 12367 F(n): 1/nlog(n) for n > 1 N: 10^{4300} F(n): 1/[nlog(n)*loglog(n)] for n > 2 N: googolplex


IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



