# Series converging

1. Jun 5, 2013

### DotKite

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

Show for what real numbers p and q $\sum \frac{1}{n^{p}ln^{q}(n)}$ diverges or converges.

2. Relevant equations

3. The attempt at a solution

I am kind of lost because it seems that with both subscripts p and q there are a bunch of cases you have to work through. My professor wasn't explicit in regards to what values p and q we had to consider.

My number 1 question would be: do you consider p and q covering the same set of numbers? Such as p>1 and q>1? Or do you have to also consider 0<p<1 while q>1 as well? It just seems this would take forever.

Anyways I think I have a solution for p,q > 1

$\frac{1}{n^{p}ln^{q}(n)}$ ≤ $\frac{1}{n^{p}}$ for all n ≥ 3

since $\frac{1}{n^{p}}$ converges (p > 1) by comparison test $\sum \frac{1}{n^{p}ln^{q}(n)}$ converges for p,q > 1

2. Jun 6, 2013

### Office_Shredder

Staff Emeritus
You can extend your comparison test more, what if q=0 and p>1 for example?

In general you are going to have to consider p and q being any number - there aren't that many actual different cases though

3. Jun 8, 2013

### DotKite

$\frac{1}{n^{p}ln^{q}(n)} \geq \frac{1}{nln(n)}$ for 0 ≤ p,q ≤ 1.

and $\frac{1}{nln(n)}$ is divergent by the integral test, thus $\frac{1}{n^{p}ln^{q}(n)}$ diverges by comparison.

Now if p,q < 0 then $\sum \frac{1}{n^{p}ln^{q}(n)}$ becomes $\sum n^{p}ln^{q}(n)$

which clearly diverges.

Ok, so all of this together we have the series converging for p,q > 1 and diverging for p,q ≤ 1. So are the other cases I need to check are ones with p and q being in these two different ranges respectively? ie p ≤ 1 and q > 1 and vice versa?

Last edited: Jun 8, 2013
4. Jun 9, 2013

### Quinzio

You saw that p,q=1 is a sort of boundary value. Then it's worthwhile to investigate it deeper.

You could check what happens if you hold still p=1 and change q, then q=1 and vary p.