For what values of p and q does a series converge?

I think you've miss typed somewhere. Your sum only has a q, not a p.

 

Hint : Try considering what's going on around ##q<0##. Try ##q=-1## for example.

How about when ##0 \le q \le 1##?

What can you conclude?

 

It's all good. I wish it was just ##q##, would've made it a bit easier.

As for ##p## and ##q## though. Do you know anything about the comparison test?

 

Hint: if [itex]q > 0[/itex] and [itex]\ln n > 1[/itex], then [itex]\displaystyle\frac{1}{n^p (\ln n)^q} < \frac{1}{n^p}[/itex].

Similarly, if [itex]p > 1[/itex] and [itex]n > 1[/itex] then [itex]\displaystyle\frac{1}{n^p (\ln n)^q} <
\frac{1}{n (\ln n)^q}[/itex].

Can you see how to integrate [itex]\displaystyle\int_2^\infty \frac{1}{x (\ln x)^q}\,\mathrm{d}x[/itex] by substitution?

 

Alright, I'll help you run through one case since there's a couple cases to consider.

The series at hand : ##\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q}##

Ask yourself, which term in the denominator of your series is your series more dependent on? That is, which function grows faster? ##n## or ##ln(n)##? Forget about ##p## and ##q## for a moment.

Without too much thought it's easy to see ##n>ln(n)## for sufficiently large ##n## independent of ##p## and ##q## and you can use that to your advantage when comparing.

##\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q} \le \sum\limits_{n=2}^\infty \frac{1}{n^p}##

So for all values of ##q##, the convergence of the series on the left depends on the convergence of the series on the right. What do you know about the series on the right? What can you conclude from all this? This will give you your first case.