Determine ## \sum \frac{1}{n^{p}ln(n)} ## diverges for 0<p<1

  • Thread starter DotKite
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  • #1
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Homework Statement



## \sum \frac{1}{n^{p}ln(n)} ##

Homework Equations





The Attempt at a Solution



All the common tests I have tried are inconclusive. I have the feeling that direct comparison test is the way to go, but I cannot seem to find what to compare ## \frac{1}{n^{p}ln(n)} ## to.
 

Answers and Replies

  • #2
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Try plugging in [itex] p=1 [/itex] and see whether that one diverges first, from there it should be easy.
 
  • #3
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i don't see how plugging p=1 gets you anywhere other than showing that it diverges for p=1. For p=1 you can use integral test.

I ended up using cauchy condensation test http://en.wikipedia.org/wiki/Cauchy_condensation_test

Then showed that the series in this test diverges (using ratio test) for 0<p<1 thus implying the original diverges
 
  • #4
Infrared
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I think kontejnjer's point is that once you establish that it diverges for p=1, you can use the comparison [itex] \frac{1}{n^pln(n)}>\frac{1}{nln(n)} [/itex] for [itex] 0<p<1 [/itex]
 
  • #5
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I think kontejnjer's point is that once you establish that it diverges for p=1, you can use the comparison [itex] \frac{1}{n^pln(n)}>\frac{1}{nln(n)} [/itex] for [itex] 0<p<1 [/itex]

ah hah. much more elegant
 
  • #6
Ray Vickson
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ah hah. much more elegant

Even more elegant: choose k > 0 so that p + k < 1. Since ##\ln(n)/n^k \to 0## as ##n \to \infty,## we have ##\ln(n) \leq n^k## for all ##n \geq N(k)##, where ##N(k)## is finite.
Thus, for ##n \geq N(k)## we have
[tex] \frac{1}{n^p \ln(n)} \geq \frac{1}{n^{p+k}} \geq \frac{1}{n} [/tex]
 

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