<s>[INST] Showing $\frac{3\sqrt{3}}{2\pi}\sum_{z\in\Lambda}\frac1{1-\left(\frac{z}{\sqrt3}-1\right)^3}=1$, with $\Lambda$ a lattice Consider the sum
$$\frac{3 \sqrt{3}}{2\pi}\sum_{z\in \Lambda} \frac{1}{1-\left(\dfrac{z}{\sqrt{3}}-1\right)^3} \overset{?}=1$$
with $\Lambda=3\mathbb Ze^{\pi i/6}+3\mathbb Ze^{-\pi i/6}$, then it is numerically not too difficult to see that this sum is equal to $1.$ I am however looking for an analytic argument of this nice fact. I assume it must rely on some subtle symmetries?
Please let me know if you have any questions.
 [/INST] Let $\Lambda' = (e^{i\pi/6}\mathbb{Z} \oplus e^{-i\pi/6}\mathbb{Z}) - \frac{1}{\sqrt{3}}$. Then by substituting $z = 3(\omega + \frac{1}{\sqrt{3}})$,
$$ S := \sum_{z \in \Lambda} \frac{1}{1-((z/\sqrt{3})-1)^3}
= \sum_{\omega \in \Lambda'} \frac{1}{1-(\sqrt{3} \omega)^3}. $$
Our goal is to show that

Claim. $\displaystyle S = \frac{2\pi}{3\sqrt{3}}$.

Now by noting that $\Lambda'$ can be decomposed into concentric "discrete regular triangles" centered at the origin, let $\Lambda'_N$ be the union of the first $N$ smallest triangles. For example, $\Lambda'_5$ is

In particular, each $\Lambda'_N$ is symmetric about a $\frac{2\pi}{3}$-rotation about the origin. Then by using the partial fraction decomposition
$$ \frac{1}{1 - w^3} = \frac{1}{3} \sum_{\xi \, : \, \xi^3 = 1} \frac{1}{1 - \xi w}, $$
we find that
\begin{align*}
S
= \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-(\sqrt{3} \omega)^3}
= \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-\sqrt{3}\omega}
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \Lambda''_N} \frac{1}{\tau},
\end{align*}
where $\tau = \frac{1}{\sqrt{3}} - \omega$ and $\Lambda''_N = \frac{1}{\sqrt{3}} - \Lambda'_N$. For example, $\Lambda''_5$ is:

Now by regrouping the dots in $\Lambda''_N$ into concentric regular triangles centered at the origin, we are left with $N-1$ triangles plus two extra "discrete lines", which we denote by $\gamma_N$. For example, the next picture demonstrates the decomposition of $\Lambda''_5$ into the triangles and $\gamma_5$:

Since the sum along each concentric triangle vanishes by symmetry, we are left with
\begin{align*}
S
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{\tau}
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{(\tau/N)} \cdot \frac{1}{N}
\end{align*}
Now the last one can be recognized as a Riemann sum for a contour integral. Indeed, if we set
$$ z_1 = -\frac{\sqrt{3}}{2} - \frac{3i}{2}, \qquad z_2 = \sqrt{3}, \qquad z_3 = -\frac{\sqrt{3}}{2} + \frac{3i}{2} $$
so that the polygonal line $\overline{z_1 z_2} \cup \overline{z_2 z_3}$ is the "limit of the rescaled discrete line $N^{-1}\gamma_N$" as $N\to\infty$, then
\begin{align*}
S
&= \frac{1}{\sqrt{3}} \int_{\overline{z_1 z_2} \cup \overline{z_1 z_2}} \frac{|\mathrm{d}z|}{z} \\
&= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \int_{[z_1, z_2]} \frac{\mathrm{d}z}{z} + \frac{1}{e^{i5\pi/6}} \int_{[z_2, z_3]} \frac{\mathrm{d}z}{z} \biggr) \\
&= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \cdot \frac{2\pi i}{3} + \frac{1}{e^{i5\pi/6}} \cdot \frac{2\pi i}{3} \biggr) \\
&= \frac{2\pi}{3\sqrt{3}}.
\end{align*}
Therefore the desired conclusion follows.
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