Without attempting to use analytic continuation on the complex plane, we will now calculate the first two values for negative odd integer values of s i.e. - 1 and - 3, for the Riemann Zeta Function.
1/(1 - s) = 1 + s + s^2 + s^3 + s^4 + s^5 + s^6 + .......
Differentiating both sides,
1/{(1 - s)^2} = 1 + 2s + 3(s^2) + 4(s^3) + 5(s^4) + 6(s^5) + .....
Setting s = - 1,
1/4 = 1 - 2 + 3 - 4 + 5 - 6 + ......
Therefore 1/4 = η(- 1)
ζ(- 1) = η(- 1)/{1 - 2^(s - 1)} = (1/4)/(- 3) = - 1/12
If we differentiate both sides again
2/{1 - s)^3 = 2 + 6s + 12(s^2) + 20(s^3) + 30(s^4) +....
Then differentiating once more,
6/(1 - s)^4 = 6 + 24s + 60(s^2) + 120(s^3) +.......
Setting s = - 1,
6/16 = 6 - 24 + 60 - 120 + .....
η(- 3) = 1 - 2^3 + 3^3 - 4^3 + 5^4 - ......
= 1 - 8 + 27 - 64 + 125 - ....
So η(- 3) + 6/16
= (1 - 8 + 27 - 64 + 125 - ....) + (6 - 24 + 60 - 120 + .....)
= 1 - 2 + 3 - 4 + 5 - ......
Thus η(- 3) + 6/16 = η(- 1)
Therefore η(- 3) = η(- 1) - 6/16
= 1/4 - 6/16 = 4/16 - 6/16 = - 2/16 = - 1/8
ζ(- 3) = η(- 3)/{1 - 2^(s - 3)} = (- 1/8)/(- 15) = 1/120
This would suggest that in principle we should be able to work out all the odd negative integer values for s through a similar process of combining already attained values for s with the varied series that arise from continued differentiation of both sides of the equation.
The question then arises as to why similar attempts with respect to even integer values of s do not hold.
As we have seen,
2/{1 - s)^3 = 2 + 6s + 12(s^2) + 20(s^3) + 30(s^4) - 42(s^5) + ....
Thus setting s = - 1,
2/8 = 2 - 6 + 12 - 20 + 30 - 42 + ....
Therefore by combining terms in successive pairs 2/8 = - 4 - 8 - 12 - ......
= - 4(1 + 2 + 3 + ....)
As the term inside the bracket is ζ(- 1), this would imply that
2/8 = - 4(- 1/12) = 1/3 which is meaningless.
However what is interesting in this case is that instead of obtaining an eta expression on the RHS (as with odd integer values for s) we obtain a zeta value.
So clearly whereas we can derive zeta values from eta, we cannot here derive a zeta value from another zeta value!
The reason is that this value for ζ(- 1) arises when we set s = + 1 on the RHS which means that the value for 2/{1 - s)^3 is thereby infinite!
Once again
when s = - 1,
2/8 = 2 - 6 + 12 - 20 + 30 - 42 + ....
η(- 1) = 1 - 2 + 3 - 4 + 5 - 6 + ....
Therefore 2/8 - η(- 1) = 1 - 4 + 9 - 16 + 25 - 36 + ...
So 2/8 - η(- 1) = η(- 2)
Therefore η(- 2) = 2/8 - 1/4 = 0.
And as ζ(- 2) = η(- 2)/{1 - 2^(s - 1)} this implies that ζ(- 2)= 0.
Thus we derive the correct answer in this case.
And a similar result in principle would emerge whenever we use an negative even integer value (i.e. - 4, - 6, -8, etc.) for s.
