What values of p make the series converge?

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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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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?
 
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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.
 
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.
 
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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).
 
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