PlanetMath: prime harmonic series

The prime ¹ 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 Erdos [2].

Proof. Assume that this series is convergent. If so, then, for a certain $k\in\mathbb{N}$ , we have:$$ \sum_{i > k}\frac{1}{p_i} < \frac{1}{2},$$ where

is the $i^{\mathrm{th}}$ prime. Now, we define $\lambda_k(n) := \char93 \{ m\leq n : p_i \div m \Rightarrow i \leq k \}$ , the number of integers less than

divisible only by the first

primes. Any of these numbers can be expressed as $m = p_1^{\omega_1} \dotsm p_k^{\omega_k} \cdot s^2,\; \omega_i\in\{0,1\}$ (i.e. a square multiplied by a square-free number). There are

ways to chose the square-free part and clearly $s\leq \sqrt{n}$ , so $\lambda_k(n) \leq 2^k \sqrt{n}$ . Now, the number of integers divisible by

less than

is $\bigl\lfloor\frac{n}{p_i}\bigr\rfloor$ , so the number of integers less than

divisible by primes bigger than

(which we shall denote by $\Lambda_k(n)$ ) is bounded above as follows:$$\Lambda_k(n) \leq \sum_{i>k}\left\lfloor\frac{n}{p_i}\right\rfloor \leq \sum_{i>k}\frac{n}{p_i} < \frac{n}{2}$$ However, by their definitions, $\lambda_k(n)+\Lambda_k(n) = n$ for all $n\in\mathbb{N}$ and so it is sufficient to find an

such that $\lambda_k(n)\leq\frac{n}{2}$ for a contradiction and, using the previous bound for $\lambda_k(n)$ , which is $\lambda_k(n) \leq 2^k \sqrt{n}$ , we see that $n = 2^{2k+2}$ works. $\qedsymbol$

The series is in some ways similar to the Harmonic series $H_n := \sum_{k = 1}^n \frac{1}{n}$ . In fact, it is well known that $H_n = \log n + \gamma + o(1)$ , where $\gamma = 0.577215664905\dots$ 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 $C = 0.26149721\dots$ and is sometimes called the Mertens constant. Its divergence, however, is extremely slow: for example, taking

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

's also slow divergence).

Bibliography

Footnotes