My Math Forum Converges Absolutely

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 December 1st, 2010, 03:58 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 $\displaystyle\sum\limits_{k=2}^{\infty} \frac{1}{log^pk}$ 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 $\frac{log^kk}{log^pk$ 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 $\emptyset$ So I am stuck. Any help would be appreciated.
 December 1st, 2010, 07:29 PM #2 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Converges Absolutely Try Cauchy's condensation test.
 December 1st, 2010, 08:11 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 p-series. P-series 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 p-series. 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, 10:08 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, 08:23 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 $\displaystyle\sum\limits_{k=2}^{\infty} \frac{1}{klog^pk}$ converges absolutely for p in (1, infinity).
December 2nd, 2010, 08:38 AM   #6
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Re: Converges Absolutely

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 Originally Posted by wannabe1 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.
I'm not sure you even need my help here! You nearly have it. (log x)^p < x for x > X (for some X), so 1/(log x)^p > 1/x for x > X. sum(x=1, infinity, 1/x) = sum(x=1, X, 1/x) + sum(x=X+1, infinity, 1/x) is infinite. But clearly sum(x=1, X, 1/x) is finite (less than X, say) so sum(x=X+1, infinity. 1/x) is infinite.

Can you take it from here?

 December 2nd, 2010, 09:04 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, 09:17 AM   #8
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Re: Converges Absolutely

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 Originally Posted by wannabe1 I am just confused, is it just a known fact that log(x)^p < x when x > X, and this works for any p?
You were the one that said that! "(logx)^p will only be larger than x for a finite number of x".

Yes, this fact is well-known.

 December 2nd, 2010, 09:39 AM #9 Senior Member   Joined: Sep 2009 Posts: 115 Thanks: 0 Re: Converges Absolutely Alright thank you very much.
 December 2nd, 2010, 03:29 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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