For what values of p does this series converge?

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Homework Help Overview

The discussion revolves around determining the values of p for which the series \(\sum_{k=2}^{\infty} \frac{1}{(\log k)^p}\) converges absolutely. The subject area includes series convergence tests and logarithmic functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants have attempted various convergence tests, including the ratio test, root test, and limit comparison test. There is a suggestion to use a comparison test with \(\frac{1}{k}\) and to consider the implications of a "p-test." However, some participants note that the series does not fit the criteria for a p-series.

Discussion Status

The discussion is active, with participants exploring different approaches to analyze the convergence of the series. Some guidance has been offered regarding comparison tests, but there is no explicit consensus on the outcome or the validity of the approaches discussed.

Contextual Notes

One participant expresses a belief that the series does not converge for any values of p, but this assertion remains unproven within the discussion. The applicability of certain tests is also questioned, particularly regarding the classification of the series.

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Homework Statement


Find all values of p for which the given series converges absolutely: [tex]\sum[/tex] from k=2 to infinity of [1/((logk)^p)].


Homework Equations




The Attempt at a Solution


I've tried the ratio test, the root test, limit comparison test ... everything. I know the answer is the null set (that is, for no values of p does the series converge), but I can't prove that rigorously.
 
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Do a comparison test with 1/k. Can you show lim k->infinity (log(k))^p/k=0?
 
have you considered simply looking at this question as a "p-test"
 
oinkbanana said:
have you considered simply looking at this question as a "p-test"
The series is not a p-series, so this test is not applicable. Here is a p-series:
[tex]\sum_{n = 1}^{\infty} \frac{1}{n^p}[/tex]
 

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