divergence of $\sum_{k_1\neq k_2\neq \dots \neq k_n}^{\infty}\frac{1}{p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}}$

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.

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).

How should we proceed to in this case?

• You don't really need very precise estimates, but if you want you can show $S_n \sim e^\gamma \log(n)$, see e.g. math.stackexchange.com/questions/1341369/… – Winther Jul 23 '17 at 14:26
• @Winther thanks! But you said You don't really need very precise estimates , how to show diverges using $S_n$ otherwise? – MAN-MADE Jul 23 '17 at 14:33
• The title seems to be a tad misleading. – Simply Beautiful Art Jul 23 '17 at 18:01
• @SimplyBeautifulArt can you please suggest one, thanks. – MAN-MADE Jul 23 '17 at 18:12
• @MANMAID How did you arrive at the calculation relating $T(n) = S(n)$? You didn't define the notation $k_1\ne k_2 \ne\cdots \ne k_n$, but it would be natural to think it means "$k_1,k_2,\ldots,k_n$ are distinct". However, the calculation suggests that you really mean "$k_1,k_2,\ldots,k_n$ are not all identical", which is a very different criterion. Of the index sets $(k_1,k_2,k_3) \in \{ (1,1,2), (1,2,1), (3,2,1) ,(1,2,3) \}$, which of these four is included in your summation? – Erick Wong Jul 23 '17 at 21:34

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 $$C_1\cdot \ln{p_n} > S_n>C_2\cdot \ln{p_n}$$
• Again can you say if $T_n$ diverges or not... yes or no at least! – MAN-MADE Jul 23 '17 at 16:05
• @MANMAID Otherwise try bounding $T_n$ below by a function of $S_n=\prod_{p \le n} \frac{1}{1-1/p}$, that $\lim_{n \to \infty} S_n$ diverges implies your series diverges – reuns Jul 24 '17 at 2:17