There's a paragraph of 104 Number Theory problems, on page $9$ that says:
From the formula $\prod_{i=1}^\infty\frac{p_i}{p_i-1} = \infty ,$
using the inequality $1+t \le e^t$, $t \in \mathbb{R}$ we can easily derive $\sum_{i=1}^{\infty}\frac{1}{p_i}= \infty.$
Here I add the two pages that contain this:
First of all, I don't understand what does that have to do with the chapter, can someone explain me that? And second, how did they arrive to the result using that inequality?
Thanks so much.
Starting: $e^{\frac{1}{p_i - 1}} \geq 1 + \dfrac{1}{p_i - 1} = \dfrac{p_i}{p_i - 1}$.
So: $e^{\displaystyle \sum_{i=1}^n \dfrac{1}{p_i - 1}} \geq \prod_{i=1}^n \dfrac{p_i}{p_i - 1}$.
So: $\displaystyle \sum_{i=1}^n \dfrac{1}{p_i - 1} \geq ln\left(\prod_{i=1}^n \dfrac{p_i}{p_i - 1}\right) \to +\infty$ as $n \to \infty$
Observe that: $p_i \geq 2$ , $\forall i \geq 1$. This means: $\dfrac{1}{p_i} \geq \dfrac{1}{2(p_i - 1)}$.
So: $\displaystyle \sum_{i=1}^n \dfrac{1}{p_i} \geq \dfrac{1}{2}\cdot \displaystyle \sum_{i=1}^n \dfrac{1}{p_i - 1} \to +\infty$ as $n \to \infty$.
Done.