How to show that $\prod_{p\text{ prime}} \frac{p}{p-1}$ diverges

Question

I was just reading a Wikipedia article regarding the existence of infinitely many primes in certain infinite arithmetic progressions, and I read something interesting- that Euler had once discovered (175) the fact that

$$\sum_{n=1}^{\infty}\frac{1}{n}=\zeta(1)=\prod_{p \text{ prime}}\frac{p}{p-1}$$

and therefore that the latter diverges to infinity. My question is as follows:

Does there exist an elementary proof that $$\prod_{p \text{ prime}}\frac{p}{p-1}$$ diverges without using the fact that $$\sum_{n=1}^{\infty}\frac{1}{n}$$ diverges?

Expecting that this problem would be on MSE already, I searched for this problem, but did not find it. Nevertheless, I would not be surprised to find that this question already does exist here in some crevasse I failed to check. In that case, of course feel free to let me know.

Answer 1

There are lots of elementary proofs of this fact, and indeed the stronger "Mertens theorem" that gives the exact asymptotic rate of divergence; there is also "Chebyshev's theorem" that gives a lower bound on the number of primes up to $x$ that's strong enough to imply this divergence via partial summation.

However, in most of these methods, some formula of the form $\sum_{n\le x} \frac1n = \log x + O(1)$ is bound to be used multiple times; and that formula is a stronger statement then the mere divergence of the harmonic series.

In the end, the divergence of $\sum_{n=1}^\infty \frac1n$ follows immediately from the elementary inequality $\sum_{n=1}^N \frac1n > \int_1^{N+1} \frac1x\,dx = \log(N+1)$; so I'm not sure that trying to avoid this fact is particularly natural.

Answer 2

$\prod_{p\leq n}\frac{p}{p-1} = (1+ 1/1)(1+1/2)(1+1/4) + .. + (1 + 1/(q-1))$ where $q$ is the largest prime $\leq n$

Consider the expression: $(1 + \frac{1}{m})^n$

In the limit if $m$ and $n$ grow at the same rate towards $\infty$ this converges to $e$. (well known).

However if we keep $m$ fixed, then as $n$ grows, this limit grows to $\infty$.

$m = 1000000, n = m, (1 + \frac{1}{m})^n \approx e $

But for same $m$, if make $n = 10m$, we will get $(1 + \frac{1}{m})^n \approx 22026.35$

We know from Euclid that there are infinite primes.

Therefore we can always take enough number of primes in the original expression to make the product larger than any positive real $M$. Hence product is divergent.