Looking for a Simple Proof of the Divergence of the Prime Harmonic Series

Question

It is well known that $$\sum_{i=1}^{\infty} \frac{1}{p_{i}}$$ diverges, where the $p_{i}$'s are the prime numbers.

Does anybody know a very elementary proof of this result that would be suitable for calculus-level students?

Many thanks.

Answer 1

There is a very nice proof (due to Erdös) that involves a minimum amount of calculus, it is the following:

If $\sum 1/p$ converges then there is an $n$ such that $$1/p_{n+1}+1/p_{n+2} + \dots < \frac{1}{2}$$ chose one $n$ with this property and consider the set $A$ of all integers composed only by the first $n$ primes: $p_1,p_2,\dots,p_n$.

let $A(x)$ be the number of elements of $A$ smaller than $x$, now you can write eny element $a \in A$ as $a=m^2s$ with $s$ squarefree and there are only $2^n$ squarefree integers made by the first $n$ primes. So for all $x$

$$ A(x) \le 2^n\sqrt{x} $$

On the other hand $x-A(x)$ counts the number of integer up to $x$ with at least one prime factor $\ge p_{n+1}$, but this number is at most $$ \left \lfloor\frac{x}{p_{n+1}} \right \rfloor + \left \lfloor\frac{x}{p_{n+2}} \right \rfloor + \dots \le \frac{x}{p_{n+1}} +\frac{x}{p_{n+2}} + \dots < \frac{x}2$$ so we get the two inequalities: $$ A(x) > \frac x2 \quad \text{and} \quad A(x) \le 2^n\sqrt{x} $$ and combining both we get for all $x$: $$ 2^{n+1} > \sqrt{x} $$ but this is clearly false if $x \ge 2^{2n+2}$ leading to a contradiction, so $\sum 1/p$ diverges.

Answer 2

Define $\zeta(s) = 1+ \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+\cdots ,$ where $s \in \mathbb{R}, s> 1$.

We know that $\zeta(s) = \prod\limits_{p \text{ prime}} \left( 1 - \dfrac{1}{p^s} \right)^{-1}.$

Taking $\log$ of both the sides, we have:

$\log \zeta(s) = -\sum\limits_{p \text{ prime}}\log \left( 1 - \dfrac{1}{p^s} \right).$

Expanding $\log$ of right-hand side, we have:

$\log \zeta(s) = \sum\limits_{p \text{ prime}} \left( \dfrac{1}{p^s} \right)+ A(s).$

The term $A(s)$ takes into account the higher powers of primes. Moreover $A(s)$ converges at $s=1$

Taking $\ \displaystyle \lim_{s\to 1}$ , the left hand side is infinity and $A(s)$ is finite, so $\sum\limits_{p \text{ prime}} \left( \dfrac{1}{p} \right)$ diverges.

Answer 3

Note: I improved my answer to show that

$\sum_{p \le n}\frac1{p} \ge \ln(\ln(n))-1 $.

Or Euler's (off the top of my head):

By unique factorization

$\begin{array}\\ \sum_{k=1}^n \dfrac1{k} &\le \dfrac1{\prod_{p \le n}(1-1/p)}\\ &=\prod_{p \le n}(1+\frac1{p-1})\\ &\le\prod_{p \le n}e^{\frac1{p-1}} \qquad\text{since } e^x \ge 1+x\\ &=e^{\sum_{p \le n}\frac1{p-1}}\\ \text{so}\\ \sum_{p \le n}\frac1{p-1} &\ge \ln(\sum_{k=1}^n \dfrac1{k})\\ &\ge \ln(\ln(n)) \qquad\text{by integral test}\\ \end{array} $

Since $\dfrac1{p-1} \le \dfrac{2}{p}$, we have $\sum_{p \le n}\frac1{p} \gt \frac12\ln(\ln(x)) $.

Here is an improvement on this last paragraph.

$\begin{array}\\ \sum_{p \le n}\frac1{p-1} -\sum_{p \le n}\frac1{p} &=\sum_{p \le n}(\frac1{p-1}-\frac1{p})\\ &=\sum_{p \le n}\frac1{p(p-1)}\\ &\le\sum_{k=2}^{n}\frac1{k(k-1)}\\ &\lt 1\\ \end{array} $

so

$\sum_{p \le n}\frac1{p} \gt \sum_{p \le n}\frac1{p-1}-1 \ge \ln(\ln(n))-1 $.