calculus - 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.

Answer

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.