In the third century BC, Euclid proved the existence of infinitely many prime numbers. In the 18th century, Leonhard Euler proved a stronger statement: the sum of the reciprocals of all prime numbers diverges to infinity. A proof by contradiction follows.

Assume that the sum of the reciprocals of the primes converges:

Define as the ith prime number. We have:

There exists a positive integer i such that:

Define N(x) as the number of positive integers n not exceeding x and not divisible by a prime other than the first i ones. Let us write this n as with k square-free (which can be done with any integer). Since there are only i primes which could divide k, there are at most choices for k. Together with the fact that there are at most possible values for m, this gives us:

The number of positive integers not exceeding x and divisible by a prime other than the first i ones is equal to x - N(x).

Since the number of integers not exceeding x and divisible by p is at most x/p, we get:

or:

But this is impossible for all x larger than (or equal to) .

Q.E.D

Here is another proof that actually gives an estimate for the sum; in particular, it shows that the sum grows at least as large as ln ln n. The proof is an adaptation of the product expansion idea of Euler. In the following, a sum or product taken over p always represents a sum or product taken over a specified set of primes.

The proof rests upon the following facts:

  • Every positive integer n can be expressed as the product of a square-free integer and a square. This gives the inequality

The product corresponds to the square-free part of n and the sum corresponds to the square part of n. (See fundamental theorem of arithmetic.)

  • The inequality

which can be obtained by considering approximating rectangles in the integral definition of ln n. (See natural logarithm.)

  • The inequality 1 + x < exp(x), which holds for all x > 0. (See exponential.)

  • The identity

Actually, the exact sum is not necessary; we just need to know that the sum converges, and this can be shown using the p-test for series. (See series.)

Combining all these facts, we see that

Dividing through by π2/6 and taking the natural log of both sides gives

∎

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