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I am looking for an $\mathbb{Z}[i]$ analogue of the alternating harmonic series: $L(1,\chi)=\sum_{n=0}^\infty (-1)^n \frac{1}{n} = \frac{\pi}{4}$.

If we try adding the reciprocals of the Gaussian integers we obtain a divergent series since there are too many terms:

$$ \sum_{(m,n) \neq (0,0)} \frac{1}{m+ni} = 0$$

If instead we take the norm $|(m+ni)|^2 = (m+ni)(m-ni) = m^2 + n^2$. Then we get the Dedekind zeta function at $1$.

$$ \zeta_{\mathbb{Z}[i]}(1) = \sum_{(m,n) \neq (0,0)} \frac{1}{m^2 + n^2} = \infty$$

This "just barely diverges" in the same way that Harmonic series gives logarithmic divergence. What if we "twist" by factors of $(-1)$?

$$ L_{\mathbb{Z}[i]}(\chi, 1) = \sum_{(m,n) \neq (0,0)} \frac{(-1)^{m+n}}{m^2 + n^2} $$

Does this series have an special value? In the case of $\mathbb{Z}$, the alternating Harmonic series is in $\mathbb{Q}\pi$, and for $\mathbb{Z}[i]$ I am guessing the value is in $\mathbb{Q}\pi^2$. Possibly warrant a separate question.

It has come to my attention the sum is not absolutely convergent.

Even if no terms are rearranged, they may be re-ordered in funny ways to give different values. In this 2-dimensional case, it may be possible that summing windows $[-M, M]\times [-N,N]$ may give different values as $M,N \to \infty$.

Another problem is that actually $\sum (-1)^n \frac{1}{n} = \log 2$ [1] and $\sum \frac{(-1)^n}{2n+1} = \sum \frac{\chi(k)}{k} = \frac{\pi}{4}$ [2]. In which case the correct analogy would be:

$$ \sum_{(m,n) \in \mathbb{Z}^2} \frac{(-1)^{m+n}}{(2m+1)^2 + (2n+1)^2} \hspace{0.25in}\text{or}\hspace{0.25in}\sum_{m+in \in \mathbb{Z}[i]} \frac{\chi(m+in)}{|m+in|^2} $$

Then I am not sure what the corresponding Dirichlet character of $\mathbb{Z}[i]$ should be [3].

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cactus314
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    Since it's not absolutely convergent, you need to say more: the result will depend on the order in which you do the sum. For example, you might ask about the limit of the sums over $[-N, N]^2 \backslash {(0,0)}$ as $N \to \infty$. – Robert Israel Jul 06 '15 at 16:39
  • You mentioned that summing over windows $[-M, M]\times [-N,N]$ might lead to a different value as $M,N \to \infty$, but I think it's not. I might add a proof of the fact if you think it's necessary. Thank you. – r9m Jul 06 '15 at 19:02
  • You have the answer using function $b_2$ formula (3) in problem n° 3 in this document – Jean Marie May 17 '21 at 08:47

1 Answers1

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Since, $m$ and $m^2$ have the same parity,

$$\sum_{(m,n) \neq (0,0)} \frac{(-1)^{m+n}}{m^2 + n^2} = \sum_{(m,n) \neq (0,0)} \frac{(-1)^{m^2+n^2}}{m^2 + n^2}$$

It boils down to this question.

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r9m
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    Note that the linked question uses a specific (reasonable) order of summation. @RobertIsrael's comment points out that OP's question doesn't specify an order of summation, and order of summation will make a difference here. – 2'5 9'2 Jul 06 '15 at 16:50
  • @alex.jordan you are right! But usually in textbook scenarios we sum either by expanding the sum on circles/square (or other compact regions) regions in $\mathbb{Z} \times \mathbb{Z}$ – r9m Jul 06 '15 at 16:54
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    Hmm, but the question linked doesn't do that. It sums over a noncompact strip, then a parallel strip, and so on. – 2'5 9'2 Jul 06 '15 at 17:00
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    @alex.jordan well we face that kind of problem with these multiple sums if the order of summation is not mentioned .. I'd guess as long as they are expanded on compact regions they converge to the same value (although I'm not sure how I would go about proving that .. I've been wanting to post a question on main about this for some time .. but before that I was planning to read around and see if there are some discussions with regard to this topic somewhere). – r9m Jul 06 '15 at 17:09