Proof that the sum of the reciprocals of the primes diverges
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:
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:
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
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