(Dan asked: What's sum{p} of 1/p prod{q<p} of 1-1/q ? ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... ... SPOILER SPACE ... Write A = 1/2 + (1-1/2) 1/3 + (1-1/2) (1-1/3) 1/5 + ... (the sum we are interested in), and then write B = (1-1/2) + (1-1/2) (1-1/3) + (1-1/2) (1-1/3) (1-1/5) + ... (a rather obvious complement thereto). Then, adding termwise, A+B = 1 + (1-1/2) + (1-1/2) (1-1/3) + ... which is obviously just 1+B. Hence A = 1. Clearly we have used nothing whatever about the primes here; the theorem in question is that for any sequence a1,a2,... (probably with some technical conditions) we have sum{i} ai prod{j<i} (1-aj) = 1. This cries out for a probabilistic interpretation, and lo there is one: consider doing something that succeeds with probability a1, and if that fails doing something that succeeds with probability a2, etc.; then, provided the a don't tend to 0 too fast, we are guaranteed to succeed eventually, and Dan's sum is precisely the probability of this, split up according to the stage at which we succeed. -- g