I intended to answer Maddle’s $p$-test question, but T. Bongers has beaten me by two minutes, so I posted my answer here to save my work.
The problem statement
This is the sum: $$\sum\limits_{n=3}^\infty\frac{1}{n\cdot\ln(n)\cdot\ln(\ln(n))^p}$$ How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.
Unpublished solution
The integral test will do.
$$
\begin{aligned}
& \int_3^{+\infty} \frac{1}{x\cdot\ln(x)\cdot\ln(\ln(x))^p} \,dx \\
&= \int_3^{+\infty} \frac{1}{\ln(x)\cdot\ln(\ln(x))^p} \,d(\ln x) \\
&= \int_3^{+\infty} \frac{1}{\ln(\ln(x))^p} \,d(\ln(\ln(x))) \\
&= \begin{cases}
[\ln(\ln(\ln(x)))]_3^{+\infty} & \text{if } p = 1 \\
\left[\dfrac{[\ln(\ln(x))}{p+1}]^{p+1} \right]_3^{+\infty} & \text{if } p \ne 1
\end{cases}
\end{aligned}
$$
When $p \ge 1$, the improper integral diverges. When $p < 1$, it converges.