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We've been told over and over $\boxed{\zeta(-1) = 1 + 2 +3 + 4 + \dots = - \frac{1}{2}}$ can be do the same over number fields?

What should be the reasonable value for the zeta function $F = \mathbb{Q}(i) = \mathbb{Q}[x]/(x^2 + 1)$:

$$ \zeta_F(-1) = \sum_{ m + in \in \mathbb{Z}[i]} \sqrt{m^2 + n^2} = 0 + 4 \Big( 1 + \sqrt{2} + 2\sqrt{3} + 2 + 2\sqrt{5} +\dots \Big)$$

Community
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cactus314
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    What you have been told over and over is wrong. – John Bentin Jul 21 '16 at 19:07
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    What we're told over and over is that the sum is $\frac{-1}{12}$, not $\frac{-1}{2}$. – hmakholm left over Monica Jul 21 '16 at 19:10
  • We know the partial sums of the infinite series don't converge, rather the infinite series defines a function that can be analytically continued and then evaluated at $-1$ and meanwhile the series evaluated at $-1$ formally looks like a given divergent series. I'd just like to pre-emptively avoid an argument where we rehash all that, as we are wont to do. – anon Jul 21 '16 at 19:13
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    Anyway, $\zeta_{\Bbb Q[i]}(s)=\zeta(s)\beta(s)$ where $\beta$ is the Dirichlet $\beta$ function. Wikipedia says $\beta$ vanishes at odd negative integers, which would make this $0$. – anon Jul 21 '16 at 19:16
  • $\zeta(-1)\ne 1+2+3+4+\ldots + n + \ldots$, that's the biggest problem. It has a finite value, but that value is not related to the given series, neither is the series you're asking about. – Adam Hughes Jul 22 '16 at 02:28

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The Dedekind zeta function $\zeta_{\mathbb{Q}(i)}$ factors as a product between the usual $\zeta$ function and a Dirichlet $L$-function $L(s,\chi_4)$ associated with the Legendre symbol $\!\!\pmod{4}$. Given the reflection formula for the $\zeta$ function, it is enough to use the reflection formula for the previous $L$-function to find the zeta-regularization of your series. Why that number should be relevant, is another question.

Jack D'Aurizio
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