*proof by contradiction*as students saw last week.

Now this result opened the doors to so many questions about the statistics of primes, many of which are still unanswered! I tried to expand on this theme of how many primes are out there.

After nearly 2000 years Euler took the next giant step in this question. Around 1737 he not only gave a different proof of Euclid's theorem, but he went way beyond that and showed that the set of primes is not too thin and in a sense there are many primes! See below for more about this. One of his statements is so intriguing when he writes

$$ \sum \frac{1}{p} =\log \log \infty $$

What he meant by this? Not so clear, but on a heuristic level one can argue that this hints at the famous

*prime number theorem*which was precisely conjectured many years later and proved even many more years later down the road in 1890's. But we are of course not yet ready to discuss all this.

To prepare for all this we decided to discuss and prove a simpler well formulated statement. More precisely we settled to prove another statement of Euler's that

$$ \sum \frac{1}{p} =\infty,$$

where the sum is over all primes of course. This statement not only shows that there are infinitely many primes, it shows that in a sense the set of primes cannot be thin! Taking a clue from the fact that the harmonic series

\( \sum \frac{1}{n} =\infty \) is divergent, let us call a subset \( S \) of integers

*fat*if \( \sum_{n\in S} \frac{1}{n} =\infty \) and

*thin*if \( \sum_{n\in S} \frac{1}{n} < \infty.\) For example the set of perfect squares is thin since we know that \( \sum \frac{1}{n^2} <\infty. \) So we set out to show that the set of all primes is a `fat' subset of the set of natural numbers. Students noticed that this is already a huge improvement over Euclid's theorem.

Euler's main tool in these types of questions related to primes was an amazing formula that he found, now called the

*Euler Product Formula*:

$$\sum_n \frac{1}{n^s}= \prod_p (1-\frac{1}{p^s})^{-1}, \quad s>1.$$

(The sum is over all natural numbers and the product is over all primes). We spent some time proving this. To do this we

1. Carefully defined infinite products,

2. Showed that in an infinite product

$$ \prod_n(1+a_n)$$

if \( \sum_n |a_n| \) is convergent then the product is convergent and in that case the product is zero iff one of its terms is zero.

3. Used unique factorization of integers into primes to prove the identity.

4. Take limits as \( s \to 1\) and use the fact that the harmonic series is divergent, to derive the divergence of the inverse primes series \(\sum \frac{1}{p}\), and hence the fact that the set of primes is not thin.

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