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

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.

• See Erdős's proof here. Jul 24, 2019 at 4:14
• @RobertIsrael. Wonderfully simple ! BTW. At the age of 18 Erdos found a new proof of Bertrand's "Postulate" (first proven in 1837) Jul 24, 2019 at 13:21

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.

• +1.... According to Wikipedia (see the link in the comment to the Q by @RobertIsrael) this is due to Erdos but there is no citation or reference. Jul 24, 2019 at 13:46
• @DanielWainfleet, I hadn't seen the comment, but I have some doubts about my attribution, I had it written in some old notes. I'll try to look later for a reference. Jul 24, 2019 at 13:58
• It's in "Proofs from The Book", which refers to "Uber die Reihe $\sum \frac{1}{p}$", Mathematica, Zutphen B 7 (1938), 1-2. Jul 24, 2019 at 23:45
• @RobertIsrael Thanks, you are right, I have edited the answer. J. Perott has a proof of Euclid's theorem with a similar flavour: If there are only $n$ primes then there are $2^n$ integers not divisible by a square so for any $N$, $$2^n > N-\sum_p N/p^2 > N(1-(\pi^2/6-1)) > N/3$$ and this is false if $N > 3\cdot 2^n$. Jul 25, 2019 at 11:04
• My edit was for typos. In the display following "On the other hand...", the subscripts on the $p$'s were (incorrectly) all $n+1.$ May 24, 2021 at 18:11

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.

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

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

• We can simplify this somewhat: (1). Each term $1/k$, for $1\le k\le n,$ occurs at least once in the complete expansion of $F(n)=\prod_{p\le n}\sum_{j=0}^{n}p^{-j}.$(Exactly once by unique factorization,but we don't need it). So $\sum_{k=1}^n(1/k)\le F(n)<\prod_{p\le n}(1-1/p)^{-1}.$ So $\prod_{p\le n}(1-1/p)\to 0.$... So if $p_j$ is the $j$th prime then $\prod_{j\in \Bbb N}(1-1/p_j)=0.$....A basic result on series is that if $a_j<1$ for all $j\in \Bbb N$ then $\sum_{j\in \Bbb N} a_j$ converges iff $\prod_{j\in \Bbb N}(1-a_j)>0.$ Jul 24, 2019 at 13:01
• A basic "companion" result is that if $a_j\ge 0$ for all $j$ then $\sum_{j\in \Bbb N}a_j<\infty$ iff $\prod_{j\in \Bbb N}(1+a_j)<\infty.$ Both of these basic results can be briefly shown, even without calculus or logarithms (and more easily with'em). Jul 24, 2019 at 13:10
• In my first comment it should say $0\le a_j<1$ instead of $a_j<1.$ Jul 24, 2019 at 13:27
• I wanted to get the rate of divergence, not just show that it diverges. I found it interesting that $\sum \frac1{p-1}$ occurred and felt good that I was able (in the addition at the end) to convert this to $\sum \frac1{p}$. Jul 24, 2019 at 15:46
• Like most of my answers, I did this off the top of me head without looking anything up, so there certainly is a chance I made a mistake. However, I just looked at it again and don't see anything wrong. Jul 24, 2019 at 22:54