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

## Homework Statement

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

## 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.

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

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
Gold Member
I think kontejnjer's point is that once you establish that it diverges for p=1, you can use the comparison $\frac{1}{n^pln(n)}>\frac{1}{nln(n)}$ for $0<p<1$

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

ah hah. much more elegant

Ray Vickson
Homework Helper
Dearly Missed
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
$$\frac{1}{n^p \ln(n)} \geq \frac{1}{n^{p+k}} \geq \frac{1}{n}$$