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

Define as the *i*th prime number. We have:

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

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