| Here's a neat trick to show that the sum of the reciprocals of the primes in fact diverges. Throughout, p will denote a prime number.
Consider instead the infinite product
P = (1-1/2)(1-1/3)(1-1/5)...(1-1/p)...
Why bother? Well, there's an elementary theorem of calculus that a product (1-a1)...(1-ak)... with ak->0 converges to a nonzero value iff the sum a1+...+ak+... converges. So instead of showing the sum 1/2+1/3+...+1/p+... converges, we'll show the product is zero.
Is our problem any easier now? Well, we can always write
1 1 1
--- = ----- * ... * ----- * ...
P 1-1/2 1-1/p
each factor of which should remind us of the formula for the geometric series! So
1/P = (1+1/2+1/4+1/8+...)(1+1/3+1/9+...)...(1+1/p+1/p2+...)...
Some standard work shows we can open the brackets (essentially because all terms are positive). The general term after opening brackets looks like 1/p1n1...pknk.
By the fundamental theorem of arithmetic, this is exactly any reciprocal of a natural number, and each natural number appears exactly once in this form!
So (after some more standard calculus work) 1/P=1+1/2+1/3+1/4+...+1/m+..., where m ranges over the natural numbers. So 1/P is the harmonic series, which diverges. In particular, P=0, and winding back further we see that the sum of the reciprocals of the primes diverges.
| 
