It is basically the same as the geometric progression............but the primes are a fraction on the bottom.............................1 over 1 - 1 over the prime...

The 1st prime is 2...............so the numerator is always 1...............1 over................1 - 1/2.........b/c the prime has an exponent of negative one.........which is just prime as the denominator...........with an s of just 1..........or negative one.....................................the 1st term is 1/1 - 1/2.......................1/1/2.........as in one over one half.............or one over 1 over two...................multiply the top and bottom by 2...............as in 2/2.............u do not change the number............b/c 2/2 is one.............do the same to the top and bottom............to get rid of the fraction on the bottom..........u get 2............b/c 1 * 2 = 2..............one top...........2 * 1/2 is 1.....................2/1 is just two...........

Euler's Product Formula

Here is an amazing formula due to Euler: SUMn=1 to infinity n-s = PRODp prime (1 - p-s)-1 . What's interesting about this formula is that it relates an expression involving all the positive integers to one involving just primes! And you can use it to prove there must be infinitely many primes. For, if there were only finitely many primes, then the right side of the expression is a finite product, and in particular for s=1. But for s=1, the left side of the equation is the harmonic series which we know must diverge! This is a contradiction, so there must be infinitely many primes. Presentation Suggestions: Interested students may wish to take a few terms on the right hand side, use a power expansion, and multiply them out... to get an idea of why the equality holds. The Math Behind the Fact: The left hand side, when s is viewed as a complex variable, is also known as the Riemann zeta function. Because of the above relationship, the study of zeta functions is closely related to the study of the