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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