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

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*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):