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-a_k)... with a_k->0 converges to a nonzero value iff the sum a₁+...+a_k+... 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/p₁^n₁...p_k^n_k. 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.