I came up with this merely because its a standard manipulation of the zeta series in its range of convergence, p > 1, to get the alternating zeta series, which converges* for p > 0: (1 - 2/2^p)*zeta(p) = 1 - 1/2^p + 1/3^p - ... But googling just now on the string bogus manipulations of alternating series , this was the top hit: "Rearranging the alternating series", at http://ecademy.agnesscott.edu/~lriddle/series/rearrang.pdf <http://ecademy.agnesscott.edu/~lriddle/series/rearrang.pdf>, which doesn't seem to include it, but may have useful references. —Dan _________________ * The resulting expression gotten by solving for zeta, along with the reflection identity for zeta: ----- xi(p) := zeta(p) gamma(p/2) / pi^(p/2) satisfies xi(p) = xi(1-p) ----- enables continuing zeta to the whole complex plane (except for poles at 0 and 1).
On Sep 12, 2015, at 6:48 AM, James Propp <jamespropp@gmail.com> wrote:
That's it! THAT'S the bogus proof Mr. Brenner showed me forty years ago!
Thanks, Dan!
Can anyone give a reference? (I assume that if a high school math teacher knew about it forty years ago, someone must have written about it in print or on the web.)
By the way, I was entering the bathroom last night, and at the moment when I put my foot on the doorstep, the idea came to me, without anything in my previous thoughts having prepared me for it, that I had in the very first paragraph of my essay made a ludicrous arithmetic error, exchanging the number of bottles with the number of drinkers, so that the story problem I proposed corresponded not to the fractions I wished to compare, but rather to their reciprocals. I did not verify this, I did not have the time for it. On returning to the bedroom I verified my hunch and fixed the error to spare myself embarrassment. :-)
Jim
On Saturday, September 12, 2015, Dan Asimov <dasimov@earthlink.net> wrote:
One very bogus analysis is that
1 - 1/2 + 1/3 - 1/4 + ...
equals the difference
(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...) - 2(1/2)(1 + 1/2 + 1/3 + . . .) _____________________________________________
= (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...)
= 0
—Dan
On Sep 11, 2015, at 12:48 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
P.S. I'm sure I've seen a slicker and easier-to-follow bogus analysis of 1 - 1/2 + 1/3 - 1/4 + ... than the one I give in the End Notes; my high school math teacher Mr. Brenner showed it to me forty years ago, but for some reason I'm unable to reconstruct it. I'm sure it's out there, in print if not on the web. Can anyone provide a lead?
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