* Gareth McCaughan <gareth.mccaughan@pobox.com> [Jan 16. 2014 08:40]:
(Dan asked: What's sum{p} of 1/p prod{q<p} of 1-1/q ?
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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
Nice. About that "technical condition": 0 = 1 - 1 = 1 - A = 1 - \sum_{n>=1}{ a_n \prod_{k=1}^{n-1}{ 1 - a_k } = \prod_{n>=1}{ 1 - a_k } (**) (last equality from 1 + \sum_{n>=1}{ a_n \prod_{k=1}^{n-1}{ 1 + a_k } } = \prod_{n>=1}{ 1 + a_k } and replacing a_k by -a_k ) Now, when is (**) actually zero?
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