What values of p make the series converge?

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Homework Help Overview

The discussion revolves around determining the positive values of p for which the series \(\Sigma_{n=2}^{\infty} \frac{1}{n( \ln (n)^{p})}\) converges. The problem is situated within the context of series convergence and involves logarithmic functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the integral test to assess convergence and raise questions about how to determine specific values of p. There is uncertainty about whether p should be treated as a positive integer or any positive real number. Some participants suggest considering Taylor approximations and substitutions to analyze the integral further.

Discussion Status

The conversation is ongoing, with participants exploring various mathematical approaches and questioning assumptions about the nature of p. Guidance has been offered regarding the integral test and Taylor approximations, but no consensus has been reached on the specific values of p that ensure convergence.

Contextual Notes

There is a mention of the assumption that p is a positive value, and some participants express confusion regarding the implications of this assumption on the convergence of the series.

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Homework Statement


Find the positive values of p for which the series converges.
[tex]\Sigma_{n=2}^{\infty} \frac{1}{n( \ln (n)^{p})}[/tex]


Homework Equations



1/n^p converges if p>1, and diverges if =<1

The Attempt at a Solution


Don't know where to begin
 
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Try an integral test.
 
Doesn't the integral tell me only whether it converges or diverges? I need to actually determine the values of p for which that will converge.
 
Whether the integral converges or diverges will depend on the value of p.
 
Is it two? Are we also assuming that p will be a positive integer rather than any positive real number?
 
What do you get if you integrate 1/(x*(ln(x))^p)? Sure, take p to be positive. If it's not then the integral definitely diverges.
 
http://texify.com/img/%5CLARGE%5C%21%5Cint_2%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7Bx%20%5Cln%28x%29%5E%7Bp%7D%7D%20%3D%20%20%20%5Cleft%5B%20%5Cfrac%7B%20%5Cln%20%28%5Cln%28x%29%5E%7Bp%2B1%7D%29%7D%7B%28p%2B1%29%7D%20%5Cright%5D_2%5E%7B%5Cinfty%7D.gif
What do I do now?
 
Last edited by a moderator:
Have you considered working with a Taylor approximation of ln(1 + n).
 
e(ho0n3 said:
Have you considered working with a Taylor approximation of ln(1 + n).

That is taught later in the chapter.
 
  • #10
razored said:
http://texify.com/img/%5CLARGE%5C%21%5Cint_2%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7Bx%20%5Cln%28x%29%5E%7Bp%7D%7D%20%3D%20%20%20%5Cleft%5B%20%5Cfrac%7B%20%5Cln%20%28%5Cln%28x%29%5E%7Bp%2B1%7D%29%7D%7B%28p%2B1%29%7D%20%5Cright%5D_2%5E%7B%5Cinfty%7D.gif
What do I do now?
What happens to ln(ln(x)^{p+1}) as x --> infty?

razored said:
That is taught later in the chapter.
OK, nevermind.
 
Last edited by a moderator:
  • #11
razored said:
http://texify.com/img/%5CLARGE%5C%21%5Cint_2%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7Bx%20%5Cln%28x%29%5E%7Bp%7D%7D%20%3D%20%20%20%5Cleft%5B%20%5Cfrac%7B%20%5Cln%20%28%5Cln%28x%29%5E%7Bp%2B1%7D%29%7D%7B%28p%2B1%29%7D%20%5Cright%5D_2%5E%7B%5Cinfty%7D.gif
What do I do now?

That's not right. Substitute u=ln(x).
 
Last edited by a moderator:

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