Prove that the logarithm family of sums diverges? Define $$a_k(n) =\frac{1}{n\log(n)\log(\log(n)) \cdots \log^{k}(n)}$$
Do all of the sums (for a fixed $k \in \mathbb{N}$):
$$A_k = \sum_{n} a_k(n)$$
Diverge?

We can use integral test. Since $\alpha_k(n)$ is decreasing in $n$, we have
$$
\sum_n \alpha_k(n) =\infty \ \ \ \Longleftrightarrow \ \ \ \int_{A}^\infty \frac{\mathrm dx}{x\cdot \log (x) \cdot\log(\log x)\cdots \log^k(x)}=\infty.
$$ Let $t_k = \log^k (x)$, $t_0 = x$. We find that $t_k = \log(t_{k-1})$ and
$$
\mathrm d t_k = \frac{\mathrm dt_{k-1}}{t_{k-1}}=\cdots=\frac{\mathrm dx}{t_{k-1}\cdots t_1 t_0}=\frac{\mathrm dx}{\log^{k-1}(x)\cdots \log(x) \cdot x}.
$$ This gives
$$
\int_{A}^\infty \frac{\mathrm dx}{x\cdot \log (x) \cdot\log(\log x)\cdots \log^k(x)}=\int_{A'}^\infty \frac{\mathrm dt_k}{t_k}=\log(t_k)\Big|^\infty_{A'}=\infty.
$$
