Showing $\sum\limits_{p \in P} \frac{1}{p}$ where $P$ is the set of all primes is divergent

Question

I've been reading proofs of this theorem, and I was wondering why the following is true:

We know that $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ diverges. I'm not sure how knowing this fact leads us to the next step:

For any $M >0$ we can find primes $p_1,\cdots p_m$ such that $\left(\sum\limits_{i=1}^{N}\frac{1}{p_1^i}\right)\cdots \left(\sum\limits_{i=1}^{N}\frac{1}{p_m^i}\right)>M.$

What I am saying is, given the opportunity, I would have never thought to look at a finite product of partial sums. I understand why everything works after the above step.

Answer 1

Note: This does not answer your question. I have changed the indices on your product. We start at $0$, not $1$. I believe that the altered indices are the "right" thing to use.

Let $M$ be any real number. Since $\sum \frac{1}{n}$ diverges, for some $N$ we have $\sum_1^N \frac{1}{n}\gt M$. Let the $p_1,p_2,\dots,p_m$ be all the primes $\le N$.

Now imagine expanding "your" (altered) product $$\left(\sum\limits_{i=0}^{N}\frac{1}{p_1^i}\right)\cdots \left(\sum\limits_{i=0}^{N}\frac{1}{p_m^i}\right).\tag{1}$$ We get all the $\frac{1}{k}$, where $k$ is a product of the form $$k=p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m},$$ where the $a_i$ range independently over all numbers from $0$ to $N$, inclusive.

These $k$ include all integers in the interval from $1$ to $N$, plus possibly more. It follows that the product (1) is greater than or equal to (actually greater) than $\sum_1^N \frac{1}{n}$, and therefore (1) is greater than $M$.

Answer 2

For $2\leq n\in \Bbb N$ let $S_n$ be the set of primes not exceeding $n.$ Let $x\in \Bbb N$ such that $2^x\geq n.$ $$\text {Let }\quad V(n)= \prod_{p\in S_n}\left(\frac {1}{1-\frac {1}{p}}\right).$$ $$\text {Then }\quad V(n) = \prod_{p\in S_n}\sum_{j=0}^{\infty}p^{-j}> \prod_{p\in S_n}\sum_{j=1}^xp^{-j}.$$ Completely expand the right-most expression above. For every $m\in \Bbb N$ with $2\leq m\leq n,$ the term $\frac {1}{m}$ will appear in the expansion as the reciprocal of a product of powers of some of the members of $S_n.$ So we have $V(n)>1+\sum_{m=2}^{n+1}\frac {1}{m}.$ Therefore $\lim_{n\to \infty}V(n)=\infty.$

Let $\pi$ be the set of all primes. Then $$\prod_{p\in \pi}(1-\frac {1}{p})=\lim_{n\to \infty}\prod_{p\in S_n}(1-\frac {1}{p})=\lim_{n\to \infty}V(n)^{-1}=0. $$

Theorem. (Elementary). If $0\leq a_j<1$ for all $j\in \Bbb N$ then $$\prod_{j=1}^{\infty}(1-a_j)=0\iff \sum_{j=1}^{\infty}a_j=\infty.$$

We have $\prod_{p\in \pi}(1-\frac {1}{p})=0.$ By the above theorem we have $\sum_{p\in \pi}\frac {1}{p}=\infty.$

This is due to Leonhard Euler.