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

1. May 30, 2013

### DotKite

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. May 30, 2013

### kontejnjer

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

3. May 30, 2013

### 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

4. May 30, 2013

### Infrared

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

5. May 30, 2013

### DotKite

ah hah. much more elegant

6. May 31, 2013

### Ray Vickson

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