# Euler's proof of infinite primes and his product formula.

From what I had read, Euler had originally proved the infinitude of primes through proving his product formula and equating the primorial to the divergent harmonic series.

$$\sum^{\infty}_{n=1}\frac{1}{n^{s}} = \prod^{\infty}_{p}\frac{1}{1-p^{-s}}$$

where p ranges through the primes. However, the proofs of this formula that I have seen all seem to rely on the fact that the sum on the left is absolutely convergent for s>1. I am sure proofs of this formula can be easily found on the web so I am not going to reproduce them here, but I found the proof which employs the sieve of Eratosthenes quite fascinating. However, that proof too relies on rearrangement and at the last step, it requires that the

$$lim_{m\rightarrow\infty}\sum^{\infty}_{m}\frac{1}{n^{s}} = 0$$

which isn't the case for the harmonic series. So is the proof actually valid for the case s=1 (or s<1) by modern standards of rigor? If not, can somebody point me towards a valid proof? I assume that the result is true even if the proofs I found were invalid as it appears to be quite celebrated.

mathman
For your last equation (m->∞), the sum is divergent for all s ≤ 1 and all m, so the limit is obviously not 0.

gb7nash
Homework Helper
If not, can somebody point me towards a valid proof?

Here's a proof that I like:

Suppose that there are a finite number of primes. So let the primes be p1, p2, ... , pn.

Now consider the integer p1p2...pn + 1. We know any integer can be decomposed into a unique factorization of primes. So:

p1p2...pn + 1 = p1e1p2e2...pnen

where ei <- N, where N includes 0. There exists a prime pi such that ei > 0. So rewriting the equation:

pi(p1p2...pi-1pi+1...pn - p1e1p2e2...pi-1ei-1piei-1pi+1ei+1...pnen) = 1, so pi divides 1, where we get the contradiction.

This is a huge abuse of notation, as i could be 1 or n. But you should get the idea.

mathman: I know, which is why is questioned the validity of the theorem.

gb7nash: I was actually looking for a proof of the Euler Product Formula, not an infinitude of primes. Thanks nevertheless, I should've been more clear.

I have read sources which state the theorem for only s > 1, and perhaps the proof and theorem does not hold for s = 1 or s < 1. So does that mean Euler's result cannot be regarded as true?

HallsofIvy
Homework Helper
It was not Euler who gave that proof of the infinitude of primes, It goes back, at least, to Euclid.

mathman
Try google "Euler's product formula". I learned 2 things (1) the expression holds for Re(s) > 1 and (2) the divergence of the series for s = 1 implies that the number of primes is infinite, i.e. the right hand side product is infinite, (so the number of primes is infinite).

Try google "Euler's product formula". I learned 2 things (1) the expression holds for Re(s) > 1 and (2) the divergence of the series for s = 1 implies that the number of primes is infinite, i.e. the right hand side product is infinite, (so the number of primes is infinite).

This is sort of what I am asking, please tell me if I am misunderstanding or missing something.

1. The infinite primes are shown through the equality of the product to the series, since the series diverges for s = 1, then the product, being equal, also diverges (hence infinite).

2. If the formula holds for Re(s)>1 then why does the series for s = 1 imply that primes are infinite? The formula has not been shown to hold for s = 1 so how can we assume the equality?

disregardthat