Analyzing the series $\sum_{n \geq 2} \frac{1}{n^p \ln^qn}$. 
Consider the series 
  $$\sum_{n \geq 2} \frac{1}{n^p \ln^qn}$$
  Prove that: 
  
  
*
  
*The series converges if $p > 1$ (and any $q$), or if $p = 1$ and $q > 1$.
  
*The series diverges if $p < 1$ (and any $q$), or if $p = 1$ and $q \leq 1$.
  

Immediately I know that we must have $p > 0$, otherwise the general term won't go to zero. 
The ratio test fails, since if we call $x_n = \frac{1}{n^p \ln^qn}$, we get that: $$\frac{x_{n+1}}{x_n} = \frac{n^p \ln^q n}{(n+1)^p \ln^q(n+1)} = \left(\frac{n}{n+1}\right)^p \left(\frac{\ln n}{\ln(n+1)}\right)^q \to 1 \cdot 1 = 1$$
Since $n^{p/n} = e^{(p \ln n)/n}$ and $\ln^{q/n}n = e^{(q \ln \ln n)/n}$, the root test also fails: $$\sqrt[n]{|x_n|} = \sqrt[n]{\frac{1}{n^p \ln^qn}} = \frac{1}{n^{p/n} \ln^{q/n}n} \to \frac{1}{1 \cdot 1} = 1$$
so, no good. I hardly think that the integral test will help here. I thought the idea was to get some condition on $p$ and $q$ by using the above tests.
Then, I'm left with comparing it with $1/n^2$ or something like it, but I'm a little lost about how to go about it. Can someone give me a hand? Thanks.
 A: With the suggestions from the comments, looking up the links given, etc, I'll post my solution. Since $\frac{1}{n^p \ln^qn}$ is decreasing and non-negative, we use Cauchy's condensation test. We have to look at: $$\sum_{n \geq 2} \frac{2^n}{(2^n)^p (\ln 2^n)^q} = \sum_{n \geq 2} \frac{1}{\ln^q2} \frac{2^n}{2^{np}n^q} = \frac{1}{\ln^q2} \sum_{n \geq 2} \frac{2^n}{2^{np}n^q} = \frac{1}{\ln^q 2} \sum_{n \geq 2} \frac{2^{n(1-p)}}{n^q}$$
Now we use the ratio test for $\sum_{n \geq 2} \frac{2^{n(1-p)}}{n^q}$. We have that:
$$\frac{2^{(n+1)(1-p)}}{(n+1)^q}\frac{n^q}{2^{n(1-p)}} = 2^{(n+1)(1-p) - n(1-p)} \left(\frac{n}{n+1}\right) \to 2^{1-p}$$
So, if $p = 1$, the test fails. If $p > 1$, $2^{1-p} < 1$, and the series converges. In the same way, $p < 1$ gives us $2^{1-p} > 1$ and the series diverges. 
Now, we only have to look at the case $p = 1$. For this, we can go back to the original series $\sum_{n \geq 2} \frac{1}{n \ln^qn}$. Now, we use the integral test. Finally, we have: $$\int_2^{+\infty} \frac{1}{x \ln^q x} \ \mathrm{d}x = \left.\frac{(\ln x)^{-q+1}}{-q+1}\right|_2^{+\infty} = \frac{1}{-q+1}\lim_{\alpha \to +\infty} \left((\ln \alpha)^{-q+1} - (\ln 2)^{-q+1}\right)$$
This is valid if $q \neq 1$. Since $(\ln 2)^{-q+1}$, we only have to look at $(\ln \alpha)^{-q+1}$. It will blow up if $-q+1 < 0$, that is, $q < 1$, so the series will diverge. On the other hand, if $q > 1$, $(\ln \alpha)^{-q+1} \to 0$ and the series will converge.
And if $p = q = 1$, we have: $$\int_2^{+\infty} \frac{1}{x \ln x} \ \mathrm{d}x = \lim_{\alpha \to +\infty} \ln(\ln \alpha) - \ln(\ln 2) = +\infty.$$
