prime harmonic series
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square-free part \PMlinkescapephraseset of primes \PMlinkescapephraseharmonic series
The prime^{1}^{1}$\mathbb{P}$ denotes the set of primes. harmonic series (also known as series of reciprocals of primes) is the infinite sum ${\sum}_{p\in \mathbb{P}}\frac{1}{p}$. The following result was originally proved by Euler (using the Euler product^{} of the Riemann Zeta function^{}) but the following extremely elegant proof is due to Paul Erdős [2].
Theorem.
The series ${\mathrm{\sum}}_{p\mathrm{\in}\mathrm{P}}\frac{\mathrm{1}}{p}$ diverges^{}.
Proof.
Assume that this series is convergent^{}. If so, then, for a certain $k\in \mathbb{N}$, we have:
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where ${p}_{i}$ is the ${i}^{\mathrm{th}}$ prime. Now, we define ${\lambda}_{k}(n):=\mathrm{\#}\{m\le n:{p}_{i}\mid m\Rightarrow i\le k\}$, the number of integers less than $n$ divisible only by the first $k$ primes. Any of these numbers can be expressed as $m={p}_{1}^{{\omega}_{1}}\mathrm{\cdots}{p}_{k}^{{\omega}_{k}}\cdot {s}^{2},{\omega}_{i}\in \{0,1\}$ (i.e. a square multiplied by a square-free number). There are ${2}^{k}$ ways to chose the square-free part and clearly $s\le \sqrt{n}$, so ${\lambda}_{k}(n)\le {2}^{k}\sqrt{n}$. Now, the number of integers divisible by ${p}_{i}$ less than $n$ is $\lfloor \frac{n}{{p}_{i}}\rfloor $, so the number of integers less than $n$ divisible by primes bigger than ${p}_{k}$ (which we shall denote by ${\mathrm{\Lambda}}_{k}(n)$) is bounded above as follows:
$$ |
However, by their definitions, ${\lambda}_{k}(n)+{\mathrm{\Lambda}}_{k}(n)=n$ for all $n\in \mathbb{N}$ and so it is sufficient to find an $n$ such that ${\lambda}_{k}(n)\le \frac{n}{2}$ for a contradiction^{} and, using the previous bound for ${\lambda}_{k}(n)$, which is ${\lambda}_{k}(n)\le {2}^{k}\sqrt{n}$, we see that $n={2}^{2k+2}$ works. ∎
The series is in some ways similar to the Harmonic series (http://planetmath.org/HarmonicSeries) ${H}_{n}:={\sum}_{k=1}^{n}\frac{1}{n}$. In fact, it is well known that ${H}_{n}=\mathrm{log}n+\gamma +o(1)$, where $\gamma =0.577215664905\mathrm{\dots}$ is Euler’s constant, and this series obeys the similar asymptotic relation^{} ${\sum}_{k=1}^{n}\frac{1}{{p}_{k}}=\mathrm{log}\mathrm{log}n+C+o(1)$, where $C=0.26149721\mathrm{\dots}$ and is sometimes called the Mertens constant. Its divergence, however, is extremely slow: for example, taking $n$ as the biggest currently known prime, the ${42}^{\mathrm{nd}}$ Mersenne prime^{} ${M}_{25964951}$, we get $\mathrm{log}\mathrm{log}({2}^{25964951}-1)+C\approx 16.9672$ (while ${H}_{{M}_{25964951}}\approx 1.79975\cdot {10}^{17}$ which is enormous considering ${H}_{n}$’s also slow divergence).
References
- 1 M. Aigner & G. M. Ziegler: Proofs from THE BOOK, 3${}^{\mathrm{rd}}$ edition (2004), Springer-Verlag, 5–6.
- 2 P. Erdős: Über die Reihe $\mathrm{\sum}\frac{\mathrm{1}}{p}$, Mathematica, Zutphen B 7 (1938).
Title | prime harmonic series |
Canonical name | PrimeHarmonicSeries |
Date of creation | 2013-03-22 15:07:16 |
Last modified on | 2013-03-22 15:07:16 |
Owner | Cosmin (8605) |
Last modified by | Cosmin (8605) |
Numerical id | 17 |
Author | Cosmin (8605) |
Entry type | Theorem^{} |
Classification | msc 40A05 |
Classification | msc 11A41 |
Synonym | series of reciprocals of primes |
Related topic | Prime |
Related topic | HarmonicSeries |
Related topic | HarmonicNumber |
Related topic | Series |
Related topic | EulersConstant |
Related topic | HarmonicSeriesOfPrimes |