DanA> 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 second thought, NO! --rwg
Okay, be that way. NON-HINT: I first saw the series sum when writing down one thing, and later noticed that it can also be seen in what seems to be a very different way. --Dan On 2014-01-15, at 2:42 PM, Bill Gosper wrote:
DanA>
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 second thought, NO!
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
(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
* 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 ?
... 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
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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Good point, Joerg. I overlooked the very common case where (assuming 0 < a_k < 1, k=1,2,3,...) oo Product (1 - a_k) > 0 n=1 . Since a_k > 0, this is true if and only if oo Sum a_k < oo, n=1 (which occurs with measure 0 in the countable-cube (0,1)^oo endowed with standard Kolmogorov product measure). --Dan On 2014-01-16, at 12:42 AM, Joerg Arndt wrote:
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?
participants (4)
-
Bill Gosper -
Dan Asimov -
Gareth McCaughan -
Joerg Arndt