Infinite sum with missing terms

Consider the following Maclaurin series :

$\begin{aligned} - \ln (1-x) & = x + \frac {x^2}2 + \frac {x^3}3 + \frac {x^4}4 + \frac {x^5}5 + \frac {x^6}6 + \cdots \\ - \ln \left(1- e^{\frac \pi 3i} \right) & = e^{\frac \pi 3i} + \frac {e^{\frac {2\pi}3i}}2 + \frac {e^{\pi i}}3 + \frac {e^{\frac {4\pi}3i}}4 + \frac {e^{\frac {5\pi}3i}}5 + \frac {e^{2\pi i}}6 + \cdots \\ & = \frac 12 + \frac {\sqrt 3}2i + \frac {-\frac 12 + \frac {\sqrt 3}2i}2 + \frac {-1+0i}3 + \frac {-\frac 12-\frac {\sqrt 3}2i}4 + \frac {\frac 12 - \frac {\sqrt 3}2i}5 + \frac {1+0i}6 + \cdots \end{aligned}$

$\begin{aligned} \implies \Im \left(-\frac 2{\sqrt 3} \ln \left(1- e^{\frac \pi 3i} \right) \right) & = 1 + \frac 12 - \frac 14 - \frac 15 + \frac 17 + \frac 18 - \frac 1{10} - \frac 1{11} + \cdots \\ & = - \frac 2{\sqrt 3}\Im \left(\ln \left(1-\frac 12 - \frac {\sqrt 3}2i \right) \right) \\ & = - \frac 2{\sqrt 3}\Im \left(\ln \left(\frac 12 - \frac {\sqrt 3}2i \right) \right) \\ & = - \frac 2{\sqrt 3}\Im \left(\ln \left(e^{-\frac \pi 3i} \right) \right) \\ & = \frac {2\pi}{3\sqrt 3} = \frac {2\pi}{3^\frac 32} \end{aligned}$

Therefore, $abc = 2 \times 3 \times \dfrac 32 = \boxed 9$ .

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Notation: $\Im(\cdot)$ denotes the imaginary part function.