Prime Consequences


Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

  • \displaystyle{\sum_{p}\frac{1}{p}}   diverges.

For proof, refer to this discussion:

  • \displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}, where s is any complex number with \text{Re}(s)>1.

The outline of proof,   when s is a real number, has been discussed here:

3 responses

  1. I’ve often wondered about what I call ‘prime decimals’ – decimals that cannot be divided by other decimals (except 0.0001, for example). I wonder if the factors of the integral part of decimal numbers affect the likelihood of those numbers being ‘prime’.


    • I think your intuition is correct, since the extra decimals won’t disturb the divisibility. For example, 0.00002 divides 0.04, since 2 divides 4 and 0.00002 < 0.04.

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  2. Pingback: Prime Consequences | Yassin Balja