# 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

complex-analysissequences-and-series

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.

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.