Series with closed-form expression [closed]

1. OP's sum is undefined. Indeed, noting that

$$ e^{i\pi/6} + e^{-i\pi/6} = \sqrt{3}, $$

the summand is undefined at $z = e^{i\pi/6} + e^{-i\pi/6}$.

2. Instead, I will compute the sum

\begin{align*} S &= \frac{27}{2\pi} \sum_{a,b\in 3\mathbb{Z}} \frac{1}{3\sqrt{3} - (a e^{i\pi/6} + be^{-i\pi/6})^3} \\ &= \frac{1}{2\pi} \sum_{a,b\in \mathbb{Z}} \frac{1}{(1/\sqrt{3})^3 - (a e^{i\pi/6} + be^{-i\pi/6})^3}. \end{align*}

as in OP's now-deleted question.

Write $\Gamma = e^{i\pi/6}\mathbb{Z}\oplus e^{-i\pi/6}\mathbb{Z}$ for the hexagonal lattice in $S$, and let $\Gamma_N$ denote the set of all lattice points in $\Gamma$ that can be reached from $0$ by at most $N$ steps. For example,

Then

$$ S = \frac{1}{2\pi} \sum_{\omega \in \Gamma} \frac{1}{(1/\sqrt{3})^3 - \omega^3} = \frac{1}{2\pi} \lim_{N\to\infty} \sum_{\omega \in \Gamma_N} \frac{1}{(1/\sqrt{3})^3 - \omega^3}. $$

By using the partial fraction decomposition

$$ \frac{1}{z^3 - \omega^3} = \frac{1}{3z^2} \left( \frac{1}{z-\omega} + \frac{1}{z - e^{2\pi i/3}\omega} + \frac{1}{z - e^{4\pi i/3}\omega} \right) $$

and the symmetry of $\Gamma_N$, it follows that

$$ S = \frac{3}{2\pi} \lim_{N\to\infty} \sum_{\omega \in \Gamma_N} \frac{1}{(1/\sqrt{3}) - \omega} = \frac{3}{2\pi} \lim_{N\to\infty} \sum_{\tau\in L_N} \frac{1}{\tau}, $$

where $\tau = \frac{1}{\sqrt{3}}-\omega$ are points in the set $L_N = \frac{1}{\sqrt{3}}-\Gamma_N$.

Now, the key observation is that most of the terms in the last sum cancel out. Indeed, the points in $L_N$ that lie in the hexagon spanned by vertices in $\Gamma_N$ are symmetric with respect to an $\frac{2\pi}{3}$-rotation about the origin:

This has to do with the fact that $\frac{1}{\sqrt{3}}$ is the center of the regular triangle with vertices $0$, $e^{i\pi/6}$, and $e^{-i\pi/6}$. So, if we write $\gamma_N$ for the points in $L_N$ that lies outside the hexagon spanned by $\Gamma_N$, then after the cancellation, we get

$$ S = \frac{3}{2\pi} \lim_{N\to\infty} \sum_{\tau\in \gamma_N} \frac{1}{\tau} = \frac{3}{2\pi} \lim_{N\to\infty} \sum_{\tau\in \gamma_N} \frac{1}{(\tau/N)} \cdot \frac{1}{N}. $$

Finally, write $z_0 = -i$, $z_1 = e^{-i\pi/6}$, $z_2 = e^{i\pi/6}$, and $z_3 = i$. Then by identifying this limit as a Riemann sum, $S$ reduces to the sum of three contour integrals:

\begin{align*} S &= \frac{3}{2\pi} \biggl( \frac{1}{z_1 - z_0} \int_{\overline{z_0 z_1}} \frac{\mathrm{d}z}{z} + \frac{1}{z_2 - z_1} \int_{\overline{z_1 z_2}} \frac{\mathrm{d}z}{z} + \frac{1}{z_3 - z_2} \int_{\overline{z_2 z_3}} \frac{\mathrm{d}z}{z} \biggr) \\ &= \frac{3}{2\pi} \biggl( \frac{i\pi/3}{z_1 - z_0} + \frac{i\pi/3}{z_2 - z_1} + \frac{i\pi/3}{z_3 - z_2} \biggr) \\ &= 1. \end{align*}