Towards Reciprocal Sum Divergence

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 \[\frac{p}{p-1}+\frac{p}{2(p-2)}+\cdots+\cfrac{p}{\left(\frac{p-1}{2}\right)\left(\frac{p-1}{2}+1\right)}.\]

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

In fact, using this result as well as the fact that there are an infinite number of prime numbers, 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.

Towards Reciprocal Sum Divergence