My solution(s): [Quoted spoiler space deleted. Unquoted spoiler space follows.] It looks as if Neil, approximately, and Gareth solved the basic problem, and Gareth found essentially the generalization I had in mind. Congratulations! ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... The first version of this series is the probability that a random positive integer has a prime factor: (*) 1/2 + 1/3 (1-1/2) + 1/5 (1-1/2)(1-1/3) + 1/7 (1-1/2)(1-1/3)(1-1/5) + ... = 1 . (Or the limit of the probability that a random positive integer <= N has a prime factor, as N -> oo.) A bit more generally: If a(n), b(n) are real for n in Z+, with 0 < a(n), b(n) < 1 and a_(n)+b(n) = 1, then because oo S = Product (a_n + b_n) = 1, n=1 it follows by expanding that S = a(1) + a(2)*b(1) + a(3)*b(2)*b(1) + a(4)*b(3)*b(2)*b(1) + ... = 1 Setting a(n) = 1/Prime[n] gives (*). --Dan --------------------------------------------------------- Gareth wrote: ----- (Dan asked: What's sum{p} of 1/p prod{q<p} of 1-1/q ? ... 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 ----- and Neil B. wrote: ----- [SPOILER] ... 1! I actually have two solutions for this, one somewhat more general (though not very rigorous) and one which is probably Dan's intended solution. The first one rearranges the sum into: (1/2)+(1-1/2)(1/3+(1-1/3)(1/5+(1-1/5)( ... ))) which can be seen as the composition of an infinite number of functions A_n[z]=1/Prime[n]+(1-1/Prime[n])*z (Bill points out this is essentially a matrix product, and that I'm just obscuring things by writing it out like this.) And now the risky bit: looking at the tail end of the infinite sum, we'll have something like A_1[A_2[ ... A_inf[A_inf[A_inf[ ... ]]]...]] Of course the only fixed point of A, and thus the fixed point of A_inf (whatever that is) is 1, and then the nested expression simplifies to 1. This can also be generalized- just replace 1/Prime[n] with any real, non-one sequence in (0,2) which converges to a non-negative number, and the sum will still come out to 1. (Caution: some of these constraints are only a few minutes old!) Second solution: Expand the entire sum (parentheses added for clarity): (1/2) + (1/3 - (1/3)*(1/2)) + (1/5 - (1/5)*(1/2)-(1/5)*(1/3) + (1/5)*(1/2)*(1/3)) +... and it's easy to see that this looks like an application of the inclusion-exclusion principle, except with multiplication instead of intersection, and asymptotic density instead of cardinality. One can carry the analogy further and claim this is the asymptotic density of the union of <composites divisible by 2>, <composites divisible by 3>, and so forth- in other words, the asymptotic density of the set of composites. Since PrimePi[x]<x/(Log[x]-4) (for x>55) and is thus less than linear, the sum must be equal to 1, although it will converge very, /very/ slowly. This could also be combined with the first solution for some set-theoretical* results which I'm almost certain have been discovered before. --Neil *number-theoretical? measure-theoretical? ----- On Wed, Jan 15, 2014 at 10:46 AM, Dan Asimov <dasimov@earthlink.net> wrote:

Okay, this has gone on entirely too long!

Shall I explain the sum

(1/2) + (1/3)(1-1/2) + (1/5)(1-1/2)(1-1/3) + (1/7)(1-1/2)(1-1/3)(1-1/5) + ...

?

Or give a hint?

--Dan

On 2014-01-10, at 2:21 PM, Bill Gosper wrote:

...

Trying for several digits of DanA's

...

Out[197]= Sum[Product[1 - 1/Prime[k], {k, 1, -1 + n}]/Prime[n], {n, 1, Infinity}]

...

(No need to special-case n=1 -- Mma does the right thing on empty products. ...

...

A tall pole in the tent!

Bill! This is a family list!

...

... Is it time to give up and beg Dan for his answer?