On 18 Jan 2014, at 02:27, LizR wrote:
The demonstration that the sum of the positive integers is -1/12
relies on the assumption that the sum of
1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 -
1 .... is 1/2
However that is by no means certain.
It is 1 - 1 + 1 - 1 + etc.
The non rigorous proof consists in adding
s = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ...
to itself, with a a lift on the right
it give s + s = 1 + (1 - 1) + (1 - 1) + (1 - 1) + .... = 1 + 0 + 0 + 0
+ 0 + 0 + 0 + 0 + ... = 1,
so 2s = 1, and s = 1/2.
That is not rigorous, but you can pursue this up to 1+2+3+4+ ... =
-1/12.
The sum could be undefined, in which case the proof simply fails. Or
it could be one of the other values mentioned - if it's 0, we get
the sum of the positive ints is either 0 or infinity (because S =
4S). If it's 1, we get the sum of the pos int = -1/6!
You allude to that funny proof of that sum. It is not rigorous at all.
The reason why 1+2+3+ = -1/12 is "accepted", in some sense, is that
the extension of the zeta function (zeta(s) = sum of 1/n^s) on the
complex plain is defined everywhere, except on 1 (= 1+0i). And that
zeta extension, on -1+0i, i.e. zeta(-1) = -1/12, and looks like
1+2+3+4+5+ ...
More direct proof are available, and make the proof you allude too,
"more rigorous", but they use special criterion of limit, different
from the usual one.
Bruno
http://iridia.ulb.ac.be/~marchal/
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