0

I am reading Fundamentals of Number Theory by LeVeque, and got stuck with proving Theorem 6.11:

$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(1-\frac{1}{p} \right)<\frac{1}{\ln{x}}, $$ where $ x \ge 2 $ and $\mathbb{P}$ is the set of prime numbers.

In the book, it is first shown that $$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right)> \ln{x}. $$ To show this, it is concluded that

$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right)>\sum_{k=1}^{x} \frac{1}{k}. \tag{$*$} $$ But this $(*)$ is precisely the part which I am unable to show/understand. I tried writing

$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right) = \text{exp} \left(\sum_{p \in \mathbb{P} \cap [2, x]} \ln \left(\frac{1}{1-1/p} \right) \right), $$ but that got me nowhere.

So how can $(*)$ be concluded?

Gary
  • 31,845
mathslover
  • 1,482
  • 4
    Expand $1/(1-1/p)$ in powers of $1/p$ using the geometric series. Perform the product and use the fundamental theorem of arithmetic. All the terms on the RHS of $(*)$ will appear plus some further terms, all positive. – Gary Mar 29 '22 at 02:33

1 Answers1

1

Just to expand on Gary's comment:

This is a classic proof idea relating two different forms of the Riemann zeta function. The idea is to expand $\frac{1}{1-1/p}$ as a geometric series. $$\prod\limits_{p\le m \text{ prime}} \frac{1}{1-1/p}= \prod\limits_{p\le m \text{ prime}} \sum\limits_{j\ge 0} \frac{1}{p^j} = \sum\limits_{j_2,j_3\ldots,j_q \ge 0} \frac{1}{2^{j_2}3^{j_3}\cdots k^{j_k}}$$ where the denominator on the right hand side has all primes less than or equal to $m$.

As the powers $j_2,\ldots,j_k$ run from $0$ upward, the denominator on the right hand side will pass through every number whose prime factors are all less than or equal to $m$. Meanwhile, in the sum $$\sum\limits_{n =1}^m \frac{1}{n},$$ the numbers in the denominator are all less than or equal to $m$, so their prime factors are as well, and therefore every term in this sum is contained in the previous sum as well.

Sasha Berger
  • 266
  • Thank you. I guess that I was confused about multiplying series by another series, but I realise now that there is nothing special in doing that. I also understand the idea of the identity $ \prod_{p \le m \ \text{prime}} \sum_{j \ge 0} \frac{1}{p^j}=\sum_{j_2, \dots, j_k \ge 0} \frac{1}{2^{j_2} \dots k^{j_k}} $, but how could I prove it rigorously? – mathslover Mar 29 '22 at 12:13