# Simplest of the many proofs the prime harmonic series diverges

Over the history of mathematics, some key facts have had multiple and different proofs developed for them. Sometimes these different proofs provide a unique insight or understanding of those facts.

For example, the infinitude of primes has an ancient proof probably due to Euclid, as well as more modern proofs such as those based on Euler's product formula, and hence the fundamental theorem of arithmetic. I'm not skilled enough to understand the additional insight that the elementary proof of the prime number theorem provided.

I am searching for the simplest proof of the fact that the prime harmonic series diverges:

$$\sum\frac{1}{p} = \infty$$

By simple, I mean requiring the least knowledge beyond secondary school mathematics.

I would be happy with a proof based on complex analysis if it is simple and elegant. I would also be happy to accept as prerequisites, if required, that the harmonic series diverges, and the Basel series converges.

The proof set out here (link) is simple but in my opinion inelegant.

I have tried to read this article by Eynden surveying several proofs and none seem simple. Erdos' proof is described as simple but I have tried several times, and failed to comprehend it.

• There’s an elementary proof in Hardy and Wright, which I found easy to follow as an undergraduate. I’m not really sure of your criterion for “simple” though. May 26, 2022 at 5:20

The simplest proof is Euler's original proof. It is based on the identity $$\sum_{n=1}^\infty n^{-s}=\prod_{p}\left(1-p^{-s}\right)^{-1}, s>1.$$ This identity is equivalent to existence and uniqueness of decomposition of integers into products of primes.

Now when $$s\to 1+$$, the LHS tends to $$\infty$$ (since the harmonic series is divergent), so the RHS must become divergent as $$s=1$$, that is $$\sum\frac{1}{p}=\infty,$$ by the elementary convergence criterion for infinite products. No complex analysis involved. Only convergence criteria for series and products with positive terms.

This is my favorite way to present Euler's proof:

By the fundamental theorem of arithmetic:

$$\prod_{p \leq N} \left(1+\frac{1}{p}+\frac{1}{p^{2}}+\cdots\right) \geq 1+\frac{1}{2}+ \cdots+\frac{1}{N}$$

for any positive integer $$N$$.

If $$N>2$$,

$$1+\frac{1}{2}+ \cdots+\frac{1}{N} > 1+\frac{1}{2}+\cdots+\frac{1}{N-1} \geq \int_{1}^{N} \frac{1}{t} \, dt = \log N.$$

Provided that $$e^{2x}>\frac{1}{1-x}$$ for any $$x \in (0, 0.5]$$, it follows that

$$\log N < \prod_{p \leq N} \left(1+\frac{1}{p}+\frac{1}{p^{2}}+\cdots\right) = \prod_{p \leq N} \frac{1}{1-\frac{1}{p}} < \prod_{p \leq N} e^{\frac{2}{p}}.$$

Therefore,

$$\sum_{p \leq N}\frac{1}{p} > \frac{\log \log N}{2}$$

for any integer $$N>2$$ and we are done.

• There is an entry on my blog on this very proof: elr3to.blogspot.com/2018/04/… Best regards! May 25, 2022 at 5:42
• Thanks Jose. This looks similar to this one: fromprimestoriemann.blogspot.com/2021/02/… I'm still searching for an even simpler one if at all possible, but yours is very good too. May 25, 2022 at 13:21
• It might look similar but there is no mention of squarefree numbers in the above proof... Further, all of the inequalities in it can be established easily. May 25, 2022 at 15:15
• thanks Jose - do you know if this is actually Euler's proof? Is it something that can be traced back to his works (which I can find in archives of his collected works?). May 26, 2022 at 9:54
• Euler's proof that $\sum 1/p$ diverges did not use the careful kinds of estimates or upper and lower bounds that are standard today in analysis. He made free use of manipulations with $\infty$. The proof appeared in his paper scholarlycommons.pacific.edu/euler-works/72. It's the very last theorem, called Theorem 19, and is on the last two pages. At the very end of the paper he writes $1/2 + 1/3 + 1/5 + 1/7 + \cdots = \ell(1 + 1/2 + 1/3 + 1/4 + 1/5 + \cdots) = \ell(\ell \infty)$, where his $\ell$ was notation for the natural logarithm.
– KCd
May 28, 2022 at 1:41