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Let $p_n$ the $n^\text{th}$ prime number. I mean $p_1=2,p_2=3,p_3=5,...$ now consider this series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{p_n}=\frac12+\frac13+\frac15+\frac17+... $

I know $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} \to \infty$ (harmonic series)

now I am looking for a simple proof to show $$\sum_{n=1}^{\infty}\frac{1}{p_n}=\frac12+\frac13+\frac15+\frac17+... \to \infty$$ thanks in advance.

Kaj Hansen
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Khosrotash
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    https://www.emis.de/classics/Erdos/textpdf/aigzieg/aigzieg.pdf ... here is the quite fantastic proof due to Erdos. – Donald Splutterwit Jul 28 '17 at 08:55
  • pages 5/6 of that book, I fully agree that this is the proof. – Henno Brandsma Jul 28 '17 at 08:57
  • @HennoBrandsma That's the point of "Proofs from THE BOOK": to compile a list of all the "the proof"'s one should know about. – Arthur Jul 28 '17 at 08:57
  • @Arthur I know, I have it at home. – Henno Brandsma Jul 28 '17 at 08:59
  • Look up Merten's Theorem (there are $3$ flavours ?) the essence is ... $\sum_{p \leq x} \frac{1}{p} =\ln ( \ln (x)) +B_1+o(1)$ ... this is probably harder? – Donald Splutterwit Jul 28 '17 at 09:08
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    @HansLundmark: Sorry. This is not a dupe, the keyword being simple. The premise of this question already assumes the answer to the other question. – Aryabhata Jul 28 '17 at 09:25
  • If you accept the fact that the primes are eventually more common than any sequence of the form $n^{1+\epsilon}$ for positive $\epsilon$ (for instance, the prime number theorem), then you can show it by comparing to the sequence $\sum1/n^{1+\epsilon}$, which is smaller, but unbounded as $\epsilon \to 0$. – Arthur Jul 28 '17 at 09:40
  • Since this has been closed, I will mention this here. Euler's proof of infinite primes can be easily changed to give us that $\sum_{p \le x} \frac{1}{p} \ge c \log \log x$ for some $c \gt 0$. It is completely elementary and quite simple. – Aryabhata Jul 28 '17 at 09:47

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