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The problem prompting this question is the following from Koblitz's Introduction to Elliptic Curves and Modular Forms 2nd ed, 1.6.1 (paraphrased):

Show the sum $\sum_{l\in \mathbb{Z}[i], \textrm{ }l\neq 0} \frac{1}{l^4}$ is a nonzero real number.

I can show the sum is a real number, but am stumped on showing it is nonzero. I can bound the sum, but this bound is not helpful. Since the sum is absolutely convergent, we can rearrange the terms in the series in the following way: \begin{align}\sum_{l\in \mathbb{Z}[i], \textrm{ }l\neq 0} \frac{1}{l^4} = 4\sum_{b=1}^\infty \frac{1}{b^4} + 4\sum_{a>0}\sum_{b>0} \frac{1}{(a+bi)^4} = 4\left(\zeta(4)+ \sum_{a>0}\sum_{b>0} \frac{1}{(a+bi)^4}\right). \end{align}

I got the right hand side by "rationalizing" the denominator to get $\frac{a^4+b^4-6a^2+b^2 +i(4ab^3-4a^3b)}{(a^2+b^2)^4}$, since the real part depends only on squares of $a,b$ we can sum over just the first quadrant and simply multiply this sum by 4. Perhaps this is the wrong approach, and there is an easier way.

More generally, I am interested in learning more about techniques which can be used to either evaluate sums over lattices or bound them (particularly from below). A resource with some worked out examples would be great, as well as some tricks you have found useful. An analogue for such a trick (for a single sum) might be something like GNUSupporter 8964民主女神 地下教會's answer to the question Prove $\lim_{m\to\infty}\sum_{k=1}^m\frac{2^{-k}}{k} = \log 2$.

While stackexchange has some questions about sums over lattices, most seem to be questions about convergence and not about techniques for evaluating them more generally.

sendit
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    The thing that occurs to me is to use the that that expression has a (convergent) Euler product expansion... Is this in the ballpark of ideas appropriate in your context? – paul garrett Mar 12 '21 at 23:04
  • The naive answer is that it is obvious it is non-zero from a numerical check (made rigorous with an estimate of the convergent rate). It is much more difficult to prove that a given series is $0$ than non-zero. But this series is also very special due to Paul Garrett's comment. – reuns Mar 12 '21 at 23:07
  • @paulgarrett Another product: $\Delta(i)$ using that $G_6(i)=0$ – reuns Mar 12 '21 at 23:15
  • @reuns, ah, yes, a classic! Great stuff! :) – paul garrett Mar 12 '21 at 23:17
  • There is also the connexion to $B(1/2,1/4)$ computing $\int_0^1 dz = \int_0^1 \frac{d\wp_i(z)}{\wp_i'(z)}=\int_C \frac{dx}{\sqrt{4 x^3+g_2(i) x}}dx$ where $C$ is a certain contour – reuns Mar 12 '21 at 23:21
  • @paulgarrett Since this series is an Eisenstein series evaluated at z=i, your comment makes me wonder if all Eisenstein series have convergent Euler products. Perhaps that is a separate question in its own right. Thank you and reuns for the instructive comments. – sendit Mar 12 '21 at 23:47
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    @C.Moore No. $G_{2k}(\tau)$ has an Euler product iff $\tau = a+b\sqrt{d}$ is a quadratic integer, $\Bbb{Z}[\tau]$ is a PID and $|\Bbb{Z}[\tau]^\times|$ divides $2k$. – reuns Mar 13 '21 at 01:28
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    One example-family of "period of Eisenstein series" that has Euler products, extending @reuns' comment wherein we depend on class-number one, consists of sums of evaluations at the (finite) set of Heegner points attached to a negative fundamental discriminant. More subtly, corresponding finite sums of integrals over geodesics corresponding to ideal classes (etc.) of real quadratic fields give Euler products. :) – paul garrett Mar 13 '21 at 02:14

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