This question already has an answer here:
It's well known that the summation over 1/p diverges just as 1/n does. However, in the case of the sum of 1/n, we can establish upper and lower bounds to the sum with the integrals over 1/n and 1/(n-1). Therefore we can say that the sum is asymptotically equal to ln(x). Can we do anything similar for the sum of the reciprocals of the prime numbers?
I suspect there isn't a neat function due to the unpredictable distribution of prime numbers.