Verify that a summation formula for x / (1+x-2x2)

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

$\frac{x}{1+x-2x^2} = \frac{1}{3} \sum_{n=1}^{\infty} \left( 1 - (-2)^n \right) x^n$

for all real $x$ such that $|x| < \frac{1}{2}$ .

First, we have

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

We recognize these as formulas for geometric series, both valid for $|x| < \frac{1}{2}$ (the first one is actually valid for $|x| < 1$ , but both are valid on the smaller interval for $x$ ),

$\begin{align*} \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \\[9pt] \frac{1}{1+2x} &= \sum_{n=0}^{\infty} (-1)^n (2x)^n. \end{align*}$

Therefore we have

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

Stumbling Robot

Verify that a summation formula for x / (1+x-2x²)

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