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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.

Lieutenant Zipp
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  • One pedantic comment that doesn't negate your main question: it is obvious that the individual products $\prod p$ and $\prod(p-1)$ diverge, since the terms don't tend to $1$. The more subtle point is that the combined product $\prod \frac{p}{p-1}$ also diverges. – Greg Martin Mar 18 '21 at 16:05
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    This is far from elementary - but a corollary of the prime number theorem is that the $n$th prime number is asymptotic to $n\log n$. And since $\sum_{n=2}^\infty \frac{1}{n\log n}$ diverges by the integral test, that implies that $\sum_{n=1}^\infty \frac{1}{p_n}$ diverges. And then, the typical proof that usually goes along with this fact (but in reverse) shows that $\sum_{n=1}^\infty \frac{1}{p_n}$ diverges implies $\prod_{n=1}^\infty \frac{p_n}{p_n - 1}$ diverges. – Daniel Schepler Mar 23 '21 at 01:34

2 Answers2

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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.

Greg Martin
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    The divergence of the harmonic series follows from the much more elementary fact that $\sum_{k=n}^{2n} \frac1k \geq \frac12.$ – Igor Rivin Mar 22 '21 at 01:43
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$\prod_{p\leq n}\frac{p}{p-1} = (1+ 1/1)(1+1/2)(1+1/4) + .. + (1 + 1/(q-1))$ 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.

IraeVid
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sku
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