# Euler's proof for the infinitude of the primes

I am trying to recast the proof of Euler for the infinitude of the primes in modern mathematical language, but am not sure how it is to be done. The statement is that:

$$\prod_{p\in P} \frac{1}{1-1/p}=\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_n\frac{1}{n}$$

Here $P$ is the set of primes. What bothers me is the second equality above which is obtained by the distributive law, applied not neccessarily finitely many times. Is that justified?

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Remember that Euler worked in the era before rigor and justifications as we do today. If it looked like a proof, and it tasted like a proof after marinating in duck sauce and grilling for an hour, then it was a proof. Or a duck. –  Asaf Karagila Apr 8 '13 at 7:05
@AsafKaragila: Is it possible to recast Euler's proof in modern form? –  Shahab Apr 8 '13 at 7:08
Yes. You could do something like limit yourself to the first $r$ primes, show that the RHS is an upper bound (as some, but not all the terms are reproduced), and then show that, as $r$ increases, every term eventually appears in the expanded sequence. –  Aaron Apr 8 '13 at 7:14
I have vague recollections that if all of the numbers ever involved are all positive, then you can make any rearrangements you like of this sort. Alas I only have vague recollections rather than memories of specific theorems. –  Hurkyl Apr 8 '13 at 7:17

## 1 Answer

It might be instructive to see the process of moving from a heuristic argument to a rigorous proof.

Probably the simplest thing to do when considering a heuristic argument involving infinite sums (or infinite products or improper integrals or other such things) is to consider its finite truncations. i.e. what can we do with the following?

$$\prod_{\substack{p \in P \\ p < B}} \frac{1}{1 - 1/p}$$

Well, we can repeat the first step easily:

$$\prod_{\substack{p \in P \\ p < B}} \frac{1}{1 - 1/p} = \prod_{\substack{p \in P \\ p < B}} \sum_{k=0}^{+\infty} \frac{1}{p^k}$$

Because all summations involved are all absolutely convergent (I think that's the condition I want? my real analysis is rusty), we can distribute:

$$\ldots = \sum_{n} \frac{1}{n}$$

where the summation is over only those integers whose prime factors are all less than $B$.

At this point, it is easy to make two very rough bounds:

$$\sum_{n=1}^{B-1} \frac{1}{n} \leq \prod_{\substack{p \in P \\ p < B}} \frac{1}{1 - 1/p} \leq \sum_{n=1}^{+\infty} \frac{1}{n}$$

And now, we can take the limit as $B \to +\infty$ and apply the squeeze theorem:

$$\prod_{\substack{p \in P}} \frac{1}{1 - 1/p} = \sum_{n=1}^{+\infty} \frac{1}{n}$$

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Thanks. I wonder if someone can clarify that absolute convergence is indeed the reason for distributivity. –  Shahab Apr 8 '13 at 7:27
Hi, if you got expressions like $a = \sum_n a_n$ and $b = \sum_n b_n$ then $a\cdot b = \sum_n( \sum_{k\leq n} a_kb_{n-k} )$. This lets you easily see why it is true for a product of expessions like $1/(1-p_1^{-1})$ and $1/(1-p_2^{-1})$. I think that you can get something similar for finite products by induction. –  André Jul 22 at 10:39