Does the following series converge: $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n\log n(\log\log n)^\alpha} $ and $\alpha>0$ ?
Using Cauchy condensation test twice:
$\begin{align} \displaystyle\sum_{n=3}^{\infty}\frac{3^n}{3^n\log 3^n(\log\log 3^n)^\alpha} &= \displaystyle\sum_{n=3}^{\infty}\frac{1}{n\log 3(\log(n\log 3)^\alpha} \\ &= \displaystyle\sum_{n=3}^{\infty}\frac{3^n}{3^n\log 3(\log(3^n\log 3)^\alpha} \\ &= \displaystyle\sum_{n=3}^{\infty}\frac{1}{\log (3) (n\log(3\log 3)^\alpha}\end {align}$
So the series converges iff $\alpha >1$ and diverges otherwise like the harmonic series.