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### Unpublished solution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.