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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 <math>p_i</math> as the ith prime number. We have:

<math> \sum_{k=1}^\infty{1\over p_{k}} = c</math>

There exists a positive integer i such that:

<math> \sum_{k=1}^\infty{1\over p_{i+k}} < {1 \over 2}</math>

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 <math>km^2</math> 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 <math>2^i</math> choices for k. Together with the fact that there are at most <math>\sqrt{x}</math> possible values for m, this gives us:

<math>N(x) \le 2^i\sqrt{x}</math>

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:

<math> x - N(x) < \sum_{k=1}^\infty{x\over p_{i+k}} < {x \over 2}</math>

or:

<math> {x \over 2} < N(x) \le 2^i\sqrt{x} </math>

But this is impossible for all x larger than (or equal to) <math> 2^{2i+2} </math>.

Q.E.D.

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