prime harmonic series

square-free part set of primes harmonic series

The prime 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:

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 $H_n=\log n+\gamma +o(1)$, where $\gamma =0.577215664905\mathrm{\dots }$ is Euler’s constant, and this series obeys the similar asymptotic relation ${\sum }_{k=1}^n\frac{1}{p_k}=\log \log n+C+o(1)$, 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 $\log \log (2^{25964951}-1)+C\approx 16.9672$ (while $H_{M_{25964951}}\approx 1.79975\cdot {10}^{17}$ which is enormous considering $H_n$’s also slow divergence).

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