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| daniel tisdale... |
 Posted: Thu Jul 31, 2008 4:40 pm |
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Sum Sum 1/i^j = 1
j=1 i=2
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In particular, Sum Sum 1/i^(2j)=3/4
and
Sum Sum 1/i^(2j+1)=1/4 where i is again from 2 to Infinity.
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Assuming my induction was correct, then the limit of the sums of Euler/Bernoulli's closed-form sums of Zeta[2k] is also 3/4, which implies the sum of Z[2k+1]=Z[3]+Z[5]+... = 1/4--remembering to sum from i=2 or subtract 1.
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I've found references for the Bernoulli algorithm--is there a paper or work that contains the limits above? I suppose these are things one comes across along the way, but I think the sum of the {closed-forms - 1) would be exhibited somewhere. |
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| daniel smith... |
| Posted: Thu Jul 31, 2008 4:44 pm |
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| and in the the first sum of course the exponents also begin at 2. |
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| daniel smith... |
| Posted: Thu Jul 31, 2008 5:05 pm |
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So my question, if it's not apparent, is: is there a reference for the proposition, easily enough proved, that
(Z[2]-1) + (z[4]-1)+(Z[6]-1)+...etc. = 3/4 ? |
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| daniel smith... |
| Posted: Thu Jul 31, 2008 6:07 pm |
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Thank you. Your 3.2.7. is my first assertion.
The assertion you (I think) question is simply that the fractional part of Sum(Zeta[2k]), that is, the fractional parts [ ] of
[Pi^2/6] + [Pi^4/90] + [Pi^6/945]...etc. = 3/4.
I thought this was interesting because of its relation to the recursive expression. I make plenty of mistakes--are you saying this does not converge to 3/4?
I will peruse your paper at leisure, thank you. |
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| Gottfried Helms... |
| Posted: Thu Jul 31, 2008 10:37 pm |
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Am 01.08.2008 05:05 schrieb daniel smith:
Quote: So my question, if it's not apparent, is: is there a reference for the proposition, easily enough proved, that
(Z[2]-1) + (z[4]-1)+(Z[6]-1)+...etc. = 3/4 ?
Perhaps you find the following interesting:
http://go.helms-net.de/math/binomial_new/10_4_zetasums.pdf
Gottfried Helms |
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| Robert Israel... |
| Posted: Thu Jul 31, 2008 11:08 pm |
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daniel smith <daniel6874 at (no spam) gmail.com> writes:
|> So my question, if it's not apparent, is: is there a reference for the
|> proposition, easily enough proved, that
|> (Z[2]-1) + (z[4]-1)+(Z[6]-1)+...etc. = 3/4 ?
It's just a matter of switching the order of summation. In Maple 12:
Quote: sum(sum(k^(-2*n),n=1..infinity),k=2..infinity);
3/4
More generally,
sum_{n=1}^infinity (zeta(2 n) - sum_{k=1}^m 1/k^(2 n))
= 1/(2 m) + 1/(2 m + 2)
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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