zeta function paradox at s=-1
- From: rokirovka@xxxxxxxxx
- Date: Mon, 3 Dec 2007 20:32:41 -0800 (PST)
The value of the zeta function at s=-1, 1+2+3+4+...., = -1/12.
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.
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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.
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.
Thus we have our answer:
1+2+4+5+7+8+10+11+.... = 3*(1/12) = 1/4.
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So what is the value of 1+2+4+5+7+8+10+11+.... ?
Is it 1/6 or 1/4 ?
Jeff Caveney
.
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