Prove that the logarithm family of sums diverges?

Question

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?

Answer 1

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 $$ \int_{A}^\infty \frac{\mathrm dx}{x\cdot \log (x) \cdot\log(\log x)\cdots \log^k(x)}=\int_{A'}^\infty \frac{\mathrm dt_k}{t_k}=\log(t_k)\Big|^\infty_{A'}=\infty. $$