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

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

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

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

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

First, we rewrite the given expression in a form we can deal with,

    \begin{align*}  \frac{x}{(1-x)(1-x^2)} &= \frac{x}{(1-x)(1-x)(1+x)} \\[9pt]  &= \frac{x}{(1-x)^2(1-x)} \\[9pt]  &= \frac{1}{2} \frac{1}{(1-x)^2} - \frac{1}{4} \frac{1}{1-x} - \frac{1}{4} \frac{1}{1+x} &(\text{partial fractions}). \end{align*}

Then, we use the geometric series formula, and we call from a previous exercise (Section 11.13, Exercise #3) that \sum nx^{n-1} = \frac{1}{(1-x)^2}, so we have

    \begin{align*}  \frac{x}{(1-x)(1-x^2)} &= \frac{1}{2} \frac{1}{(1-x)^2} - \frac{1}{4} \frac{1}{1-x} - \frac{1}{4} \frac{1}{1+x} \\[9pt]  &= \frac{1}{2} \sum_{n=0}^{\infty} nx^{n-1} - \frac{1}{4} \sum_{n=0}^{\infty} x^n - \frac{1}{4} \sum_{n=0}^{\infty} (-1)^n x^n \\[9pt]  &= \frac{1}{2} \left( \sum_{n=0}^{\infty} nx^n - \frac{1}{2} \sum_{n=0}^{\infty} (1 - (-1)^n)x^n \right) \\[9pt]  &= \frac{1}{2} \left( \sum_{n=1}^{\infty} nx^n - \frac{1}{2} \sum_{n=1}^{\infty} (1 - (-1)^n) x^n \right) \\[9pt]  &= \frac{1}{2} \sum_{n=1}^{\infty} \left( n - \frac{1 - (-1)^n}{2} \right) x^n. \end{align*}

This valid for all real |x| < 1 since that was where the geometric series formulas we used were valid.

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