On Sunday 30 March 2008, Steve Witham wrote:

oo ------ | | 1 | | 1 - ----- | | f(i) i = 1

f(i) = i + 1 (1/2) (2/3) (3/4) (4/5)... = 0

f(i) = (i+1)^2 (1/4) (8/9) (15/16) (24/25) ... = 0

Is the result always zero when f(i) is a positive polynomial of i?

product(1-a) diverges to 0 iff sum(a) diverges to oo. Your second product isn't in fact 0. It's 3/4 . 8/9 . 15/16 . 24/25 . ... = (1.3)/(2.2) (2.4)/(3.3) (3.5) / (4.4) ... = 1/2 (other factors cancel).

f(i) = i'th prime (1/2) (2/3) (4/5) (6/7) (10/11) (12/13) ... = 0

(The fraction of counting numbers that aren't divisible by any prime.) Does that equation have a name?

product (1-p^-s) = 1 / sum (n^-s) = 1/zeta(s) and indeed the zeta function has a pole at 0 so the product is 0.

f(i) = 2^i (1/2) (3/4) (7/8) (15/16) ... ~= 0.288788095087

Is the result always > 0 when f(i) = k^i and k > 1?

Yes (see above).

If so, does this put the primes between polynomials and exponentials?

No, it puts them between first-degree and second-degree polynomials. -- g