December 1st, 2010, 04:58 PM  #1 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Converges Absolutely
So we need to find all the values of p in R for which the given series converges absolutely. So the series is So I have tried several different tests, but I always get stuck. So the integral test will not work, since it will be hard to integrate this function. I tried the root test but I just get And the limit as k goes to infinity of the top part is infinity. For the bottom part we cannot have p in (infinity, 1) or else this will go to zero. And p cannot be in the rest since this will not be well defined. I tried the ratio test and then tried using l'hopital's rule but that did not help either. I would have used the p test and thus this works for p in (1, infinity) but I know that the answer is said to be So I am stuck. Any help would be appreciated. 
December 1st, 2010, 08:29 PM  #2 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: Converges Absolutely
Try Cauchy's condensation test.

December 1st, 2010, 09:11 PM  #3 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Converges Absolutely
Firstly, I agree with dman315. Cauchy's condensation test makes this a very easy question. Secondly, you misused pseries. Pseries only work for 1/x^p, not 1/(logx)^p. Saying for a moment that I don't know Cauchy Condensation, it seems most natural to me to simply do a basic comparison with a pseries. For that matter, do a basic comparison with the harmonic series. (logx)^p will only be larger than x for a finite number of x. 
December 1st, 2010, 11:08 PM  #4 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Re: Converges Absolutely
So I have not learned Cauchy's condensation test yet. So for comparing it to the harmonic series, I am not quite sure how to do this. I do not get this: (logx)^p will only be larger than x for a finite number of x, since we do not know what p is. 
December 2nd, 2010, 09:23 AM  #5 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Re: Converges Absolutely
If it helps any, in the problem before I proved that converges absolutely for p in (1, infinity).

December 2nd, 2010, 09:38 AM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Converges Absolutely Quote:
Can you take it from here?  
December 2nd, 2010, 10:04 AM  #7 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Re: Converges Absolutely
So I mean I understand that since it is greater than a divergent sequence then itself will diverge also by the comparison test. That part makes sense. I am just confused, is it just a known fact that log(x)^p < x when x > X, and this works for any p? Thank you for your help so far.

December 2nd, 2010, 10:17 AM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Converges Absolutely Quote:
Yes, this fact is wellknown.  
December 2nd, 2010, 10:39 AM  #9 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Re: Converges Absolutely
Alright thank you very much.

December 2nd, 2010, 04:29 PM  #10 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Converges Absolutely
If you wanted to show this, i.e. that log^p (x) will be less than x for only finitely many x, take the limit of x/log^p(x) as x goes to infinity. You know L'Hopital, I presume. This indicates that x will eventually outgrow logx, always.


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