coverges or not. Well, the answer is: It does not converge but grow beyond any limit. The following proof follows ideas by the famous Paul Erdös, and is an extended and translated version taken from www.mathepedia.de.

We numerate the prime numbers in increasing order, i.e

Assume that the harmonic series of primes would converge. Then there would be a number m such that

and consequently, for any number N (which we will choose below).

Let us define now the "small prime numbers" to be {p

_{1}, p

_{2}, ...., p

_{m}} and the "large prime numbers" to be all others.

Further, let NS be the count of numbers ≤ N that have only small prime factors (in the above sense) and let NL be the count of numbers ≤ N that have at least one large prime factor. Thus

NL + NS = N

We observe that [N/p] (with [x] denoting the integer part of x) is the number of multiples of p smaller or equal N and therefore

To estimate NS, we write any number n that is made up of only small prime factors as

where a

_{n}is the product of all square free prime factors and b_{n}^{2}be the product of quadratically occuring prime factors. (An example: for n=2400, a_{n}would be 6, b_{n}would be 20).
In a

_{n}, any small prime number occurs either with an exponent of 1 or 0. That is, there are at most 2^{m}possibilities for a_{n}. For b_{n}, we obtain b_{n}≤ √n ≤ √ N that there are at most √ N possibilities.
Therefore

NS ≤ 2

^{m}√ N
on the other hand, due to NL< N/2, we have NS > N/2.

When we choose N (which is still free to be chosen) as N= 2

^{2m+2}, we obtain that
NS ≤ 2

^{2m+1}< NS ,
a contradiction. Therefore, the assumption "The series of reciprocal primes converges" must have been wrong.