Let $p_1=2<p_2<p_3<\dots$ are consecutive prime numbers.
Let $$S_n=\sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\dots \sum_{k_n=0}^{\infty}\frac{1}{p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}}$$
It is easy to check that $S_1=2,S_3=3$ and in general $S_n=\frac{1}{1-\frac{1}{p_1}}\frac{1}{1-\frac{1}{p_2}}\dots \frac{1}{1-\frac{1}{p_n}}$
Then if we can show that as $n\to\infty$, $S_n$ diverges to infinity we can say $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges to infinity.
If we can show that $S_n\thicksim f(n)$, where $f(n)$ increases to infinity as $n\to\infty$, then we are done.
What is this $f(n)$?
Is there any other way to show $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges using $S_n$?
But I have no idea about the following problem:
$$T_n=\sum_{k_1\neq k_2\neq \dots \neq k_n}^{\infty}\frac{1}{p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}}$$
Here $k_1\neq k_2\neq \dots\neq k_n$ means they are distinct($n!$ values).
A hint: since $$S_n=\frac{p_1}{p_1-1}\frac{p_2}{p_2-1}...\frac{p_n}{p_n-1}$$
From Mertens' 3rd theorem $$0<(1-\varepsilon)e^{-\gamma}<\ln{p_n} \cdot \prod_{k=1}^{n}\frac{p_k-1}{p_k} < (1+\varepsilon)e^{-\gamma}$$ or $$\frac{\ln{p_n}}{(1-\varepsilon)e^{-\gamma}}>\prod_{k=1}^{n}\frac{p_k}{p_k-1} > \frac{\ln{p_n}}{(1+\varepsilon)e^{-\gamma}}$$ Altogether $$C_1\cdot \ln{p_n} > S_n>C_2\cdot \ln{p_n}$$