## Prime Numbers and Harmonic Series

## The Statement

There is a prime number related result which is adapted from a tutorial question in a course about number theory, which is stated as follows.

For any prime $p$, $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}$ is divisible by $p$.

### The Proof

A proof of this result is fairly straightforward, when observed correctly.

Indeed, when $p=2$, $1+\frac{1}{p-1}=1+1=2$ is divisible by $2$.

Now, we only consider the odd primes, then we should have an even number of terms for the sum above. Now here's the key part: **note that the terms of the sum above can be rearranged such that the first and last term are next to each other, as well as the second and second last term, and so on**. When we sum each of the pairs together, we then obtain

We can then see that we are able to factorise $p$ out of the sum, so it is indeed divisible by $p$.

In fact, using this result as well as the fact that there are an infinite number of prime numbers, which makes the set of prime numbers to be unbounded above due to the definition of prime numbers itself that necessitates them to be integers, it follows that the series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges to positive infinity. This provides an alternative proof to the ingenious method that produces an infinite sum of one halves.

## How about the other way around?

This gives rise to another interesting question: can we prove that there are infinitely many primes, assuming the divergence of the harmonic series?

In fact, this is true. If we break down the denominator of each term of the harmonic series into its respective prime factorisation, then apply the distributive law and the formula for the geometric infinite sum, we then obtain the following result due to Euler, where $\mathbb{P}$ denotes the set of prime numbers:

$\sum _{i=1}^{\infty }{\frac {1}{i}}=\prod _{p\in \mathbb {P} }\left(1+{\frac {1}{p}}+{\frac {1}{p^{2}}}+\cdots \right)=\prod _{p\in \mathbb {P} }{\frac {1}{1-1/p}}.$What Euler did next was taking logarithms on both sides and applying the Taylor series of logarithms gives the following:

$\displaystyle \ln \prod _{p\in \mathbb {P} }{\frac {1}{1-1/p}}=\sum _{p\in \mathbb {P} }\ln {\frac {1}{1-1/p}}=\sum _{p\in \mathbb {P} }\left({\frac {1}{p}}+{\frac {1}{2p^{2}}}+{\frac {1}{3p^{3}}}+\cdots \right)=\sum _{p\in \mathbb {P} }{\frac {1}{p}}+\text{convergent terms}.$Since this is equal to the divergent harmonic series and a finite sum containing finite terms must converge, it follows that there must be infinitely many primes. This result also implies that the sum of reciprocals of prime numbers diverges, which was more rigorously verfied by Franz Mertens.