How to prove that the partial Euler product of primes less than or equal x is bounded from below by log(x)? [closed]

How does one prove $\prod_{p \leq x}(1 - \frac{1}{p})^{-1} \geq \log(x)$?

Solution 1:

As commented by Daniel Fischer, I think you mean

$\prod_{p \leq x}\frac1{1 - \frac{1}{p}} \geq \log(x) $.

Let $P(x)$ be the set of positive integers with all of their prime factors $\le x$. Every positive integer $\le x$ is obviously in $P(x)$. Therefore,

$\begin{align} \prod_{p \leq x}\frac1{1 - \frac{1}{p}} &=\prod_{p \leq x}\sum_{k=0}^{\infty} \frac1{p^k}\\ &=\sum_{n \in P(x)} \frac1{n}\\ &>\sum_{1 \le n \le x} \frac1{n}\\ &> \ln x \end{align} $

by one of the usual harmonic sum bounds.

Note: As in many of my answers, nothing here is original.