Finding the $p,\ r,\ q$ for which the series converges I'm dealing with the series: $$\sum_{n=3}^{\infty} \frac{1}{n^p(\ln n)^q(\ln(\ln n))^r},$$ looking for the set of all $p,q,r$ such that the series converges.  Is there a way to determine this without use of the integral test?  In that case I would substitute $u=\ln x$ and so on, but I'm wondering if there is a method using ratio test, root test, comparison test, condensation criteria etc.  Any help is appreciated.
 A: First, suppose $p=q=1$. We can then use the integral test. 
If $r=1$ we have
$$
\int_3^{y} \frac{dx}{x \ln(x) (\ln(\ln(x)))} = \left[\ln(\ln(\ln(x)))\right]_3^y
=\ln(\ln(\ln(y))) -\ln(\ln(\ln(3)))
$$
which diverges as $y \to \infty$. For $r \neq 1$ we have
$$
\int_3^{y} \frac{dx}{x \ln(x) (\ln(\ln(x)))^r} = \frac{1}{1-r}\left[\ln(\ln(x))^{1-r}\right]_3^y = \frac{1}{1-r}\left(\ln(\ln(y))^{1-r} - \ln(\ln(3))^{1-r}\right)
$$
which diverges as $y \to \infty$ if and only if $r < 1$.  So when $p = q =1$, the series converges when $ r>1$ and diverges when $r \le 1$.
When $p \neq 1$, we can use the comparison test. If $a_n = \frac{1}{n^p (\ln(n))^q (\ln(\ln(n)))^r}$ with $p < 1$, then let $b_n = \frac{1}{n (\ln(n)) (\ln(\ln(n)))}$. Then
$$
\frac{a_n}{b_n} = \frac{n (\ln(n)) (\ln(\ln(n)))}{n^p (\ln(n))^q (\ln(\ln(n)))^r} = n^{1-p} (\ln(n))^{1-q} (\ln(\ln(n)))^{1-r}
$$
Since $1-p$ is positive, $n^{1-p}$ goes to infinity as $n$ goes to infinity, and will dominate the other two terms, so the ratio $\frac{a_n}{b_n}$ goes to infinity.  Since $\sum_3^\infty b_n$ diverges by the integral test, so does $\sum_3^\infty a_n$ by the comparison test.
Similarly, if $a_n = \frac{1}{n^p (\ln(n))^q (\ln(\ln(n)))^r}$ with $p > 1$, then let $b_n = \frac{1}{n (\ln(n)) (\ln(\ln(n)))^2}$. Then
$$
\frac{a_n}{b_n} = \frac{n (\ln(n)) (\ln(\ln(n)))^2}{n^p (\ln(n))^q (\ln(\ln(n)))^r} = n^{1-p} (\ln(n))^{1-q} (\ln(\ln(n)))^{2-r}
$$
Since $1-p$ is negative, $n^{1-p}$ goes to zero as $n$ goes to infinity, and will dominate the other two terms, so the ratio $\frac{a_n}{b_n}$ goes to zero.  Since $\sum_3^\infty b_n$ converges by the integral test, so does $\sum_3^\infty a_n$ by the comparison test.
A similar argument shows that the series converges when $p=1, q > 1$ and diverges when $p=1, q < 1$.
Putting it all together, the series converges if either:
$p>1$
$p=1, q > 1$
$p=q=1, r > 1$
and diverges otherwise.
A: This is an existence problem, so really we just need to find a $p$, $q$, and $r$ such that the series converges. Well, suppose that $p = q = 1$ and $r = 2$. Then, assuming that the integral test conditions are met for $\displaystyle f(x) = \frac{1}{x\ln(x)(\ln(\ln(x)))^2}$, then
$\begin{aligned}[t]\int_3^\infty \frac{1}{x\ln(x)(\ln(\ln(x)))^2} & = -\lim_{b \to \infty} \left[\frac{1}{\ln(\ln(x))}\right]_3^b \\
& = -\lim_{b \to \infty}\left[\frac{1}{\ln(\ln(b))} - \frac{1}{\ln(\ln(3))}\right] \\
& = -\left[0 - \frac{1}{\ln(\ln(3))}\right] \\
& = \frac{1}{\ln(\ln(3))}\end{aligned}$.
By the integral test (again, assuming the conditions are met which they do), since the integral of $f(x)$ converged, then you may assume the series converges also.
EDIT: It's also important that you show that all three conditions are met for the integral test for all $n \geq 3$. That is left to you to verify on $f(x)$.
EDIT2: I see that you've edited the question to say the set of all $p$, $q$, and $r$ such that the series converges. It is then not an existence problem. Using the integral test in this manner will only work for $(p,q,r) = (1,1,t)$ where $t > 1$.
