Re: zeta function paradox at s=-1

rokirovka@xxxxxxxxx writes:

The value of the zeta function at s=-1, 1+2+3+4+...., = -1/12.

Yes, zeta(-1) = -1/12, but it's more than a bit of a stretch to call that
1+2+3+4+....

Here is a paradox: what is the value of the infinite series
1+2+4+5+7+8+10+11+.... ?

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Answer #1:

Begin with 1+2+3+4+.... = -1/12.

Multiply both sides by 3:
3*(1+2+3+4+....) = 3*(-1/12)
3+6+9+12+.... = -1/4.

Subtract that last infinite series from the one we began with:
(1+2+3+4+....) - (3+6+9+12+....) = -1/12 - -1/4

And we have our answer:
1+2+4+5+7+8+10+11+.... = 1/6.


OK, if f(s) = 1^(-s) + 2^(-s) + 4^(-s) + 5^(-s) + 7^(-s) + 8^(-s) + ...
then for Re(s) < -1, f(s) = (1 - 3^(-s)) zeta(s), and the analytic
continuation to s=-1 is 1/6.

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Answer #2:

Begin with 1+2+3+4+.... = -1/12.

Multiply both sides by 2:
2*(1+2+3+4+....) = 2*(-1/12)
2+4+6+8+.... = -1/6.
Subtract the last infinite series from the one we began with:
(1+2+3+4+....) - (2+4+6+8+....) = -1/12 - -1/6
1+3+5+7+.... = 1/12.
We can express this series as Sum(n=1 to infinity) 2n-1.

OK, sum_{n=1}^infty (2n-1)^(-s) = (1-2^(-s)) zeta(s),
and at s=-1 this is 1/12.

Now take the series 1+2+4+5+7+8+10+11+....
We can express this series as Sum(n=1 to infinity) (3n-2)+(3n-1).
(3n-2)+(3n-1) = 6n-3 = 3(2n-1).
Therefore this series equals 3 times the above-mentioned series,
Sum(n=1 to infinity) 2n-1.

This doesn't fit with the zeta function interpretation.
sum_{n=1}^infty (3n-2)^(-s) + sum_{n=1}^infty (3n-1)^(-s)
is not in general sum_{n=1}^infty (6n-3)^(-s).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.