(open) Divergent series of inverse primes

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SUMMARY

The discussion centers on proving by contradiction that the series of the reciprocals of all primes diverges, specifically represented as $$\sum_{p\in \mathbb{P}}\dfrac{1}{p} =\sum_{p\;\text{prime}}\dfrac{1}{p}$$. This conclusion is established through the properties of prime numbers and their distribution. An immediate corollary of this result is the divergence of the harmonic series, which is foundational in number theory.

PREREQUISITES
  • Understanding of prime numbers and their properties
  • Familiarity with series and convergence concepts in mathematics
  • Knowledge of proof techniques, particularly proof by contradiction
  • Basic grasp of number theory and harmonic series
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  • Study the properties of prime numbers and their distribution
  • Learn about the divergence of the harmonic series
  • Explore advanced proof techniques in mathematics, especially proof by contradiction
  • Investigate the implications of the Prime Number Theorem
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Mathematicians, students of number theory, and anyone interested in the properties of prime numbers and series convergence.

fresh_42
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Show by contradiction that
$$
\sum_{p\in \mathbb{P}}\dfrac{1}{p} =\sum_{p\;\text{prime}}\dfrac{1}{p}
$$
diverges. Which famous result is an immediate corollary?

Skeleton (proof by Pál Erdös):
  1. $$\underbrace{p_1<\ldots <p_k<}_{\text{small primes}}\underbrace{\underbrace{p_{k+1}<\ldots}_{\text{big primes}}}_{\displaystyle{\sum_{j>k}\dfrac{1}{p_j}<\dfrac{1}{2}}}$$
  2. Set ##N_b =\# \{n \leq N\,|\,\exists \,j>k\, : \,p_j|n\}## and ##N_s=\#\{n\leq N\,|\,p_j|n\Longrightarrow j\leq k\}.##
  3. ##N_b < \dfrac{N}{2}##
  4. ##N_s < 2^k\sqrt{N}##
  5. ##N_b+N_s < N##
 
Last edited:
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