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

[DotKite]

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.

 

[kontejnjer]

Try plugging in p=1 and see whether that one diverges first, from there it should be easy.

 

[DotKite]

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

 

[Infrared]

I think kontejnjer's point is that once you establish that it diverges for p=1, you can use the comparison \frac{1}{n^pln(n)}&gt;\frac{1}{nln(n)} for 0&lt;p&lt;1

 

[DotKite]

I think kontejnjer's point is that once you establish that it diverges for p=1, you can use the comparison \frac{1}{n^pln(n)}&gt;\frac{1}{nln(n)} for 0&lt;p&lt;1

ah hah. much more elegant

 

[Ray Vickson]

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
\frac{1}{n^p \ln(n)} \geq \frac{1}{n^{p+k}} \geq \frac{1}{n}