[math-fun] Fraction of indivisible numbers

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