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-

`a`_{1})...(1-

`a`_{k})... with

`a`_{k}->0

converges to a

nonzero value

iff the sum

`a`_{1}+...+

`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/`p`^{2}+...)...

Some standard work shows we can open the brackets (essentially because all terms are positive). The general term after opening brackets looks like 1/`p`_{1}^{n1}...`p`_{k}^{nk}.
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.