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Verify a summation formula for 1 / (x2 + x + 1)

by RoRi

Assume that \frac{1}{x^2+x+1} has a power-series representation in terms of powers of x, verify that it has the form

    \[ \frac{1}{x^2+x+1} = \frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \sin \left(\frac{2 \pi (n+1)}{3}\right)x^n \]

valid for all real x such that |x| < 1.

First, we rewrite the given expression

    \begin{align*} \frac{1}{x^2+x+1} &= \frac{x-1}{x^3-1} \\[9pt] &= \frac{1-x}{1-x^3} \\[9pt] &= (1-x) \cdot \left( \frac{1}{1-x^3} \right). \end{align*}

Using the geometric series we have

    \[ \frac{1}{1-x^3} = \sum_{n=0}^{\infty} (x)^{3n}. \]

Therefore we have

    \begin{align*} &\frac{1}{x^2+x+1} = (1-x) \cdot \sum_{n=0}^{\infty} (x)^{3n} \\[9pt] &= \sum_{n=0}^{\infty} x^{3n} - \sum_{n=0}^{\infty} x^{3n+1} \\[9pt] &= \frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \frac{\sqrt{3}}{2} x^{3n} + \frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \left( \frac{-\sqrt{3}}{2} \right) x^{3n+1} + \frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} (0) x^{3n+2} \\[9pt] &= \frac{2}{\sqrt{3}} \left( \sum_{n=0}^{\infty} \sin \left( \frac{2 \pi}{3} \right) x^{3n} + \sum_{n=0}^{\infty} \sin \left( \frac{2 \cdot 2 \pi}{3} \right) x^{3n+1} + \sum_{n=0}^{\infty} \sin \left( \frac{3 \cdot 2 \pi}{3} \right)x^{3n+2} \right) \\[9pt] &= \frac{2}{\sqrt{3}} \left( \sum_{n=0}^{\infty} \sin \left( \frac{2 \pi (3n+1)}{3} \right)x^{3n} + \sum_{n=0}^{\infty} \left( \frac{2 \pi (3n+2)}{3} \right) x^{3n+1} + \sum_{n=0}^{\infty} \sin \left( \frac{2 \pi (3n+3)}{3} \right)x^{3n+2} \right) \\[9pt] &= \frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \sin \left( \frac{ 2 \pi (n+1)}{3} \right) x^{3n}. \end{align*}

This was the requested identity.

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