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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: enter image description here enter image description here

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.

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  • $\begingroup$ It's a logarithmic trick, $\log (ab) = \log (a) + \log (b) $. Use this with the inequality given. $\endgroup$ – Jeb May 12 '14 at 2:26
  • $\begingroup$ I added the two pages that "explain" this. Do you know what does this have to do with the previous stated? $\endgroup$ – user149455 May 12 '14 at 2:28
  • $\begingroup$ The first page basically shows there are infinitely many primes via factorization with the Harmonic series. The second page shows that the "spacing" between the primes is like the Harmonic series. $\endgroup$ – Jeb May 12 '14 at 2:31
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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.

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