prime harmonic series

square-free partset of primesharmonic 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 ${\lambda }_k(n):=\mathrm{\#}\{m\le n:p_i\mid m\Rightarrow i\le k\}$, thenumber 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).