Question

How do you find a power series representation for `x/(1-x^2)` and what is the radius of convergence?

Answer 1

Use the Maclaurin series for `1/(1-t)` and substitution to find:

`x/(1-x^2) = sum_(n=0)^oo x^(2n+1)`

with radius of convergence `1`.

Explanation:

The Maclaurin series for `1/(1-t)` is `sum_(n=0)^oo t^n`

since `(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1`

Substitute `t = x^2` to get:

`1/(1-x^2) = sum_(n=0)^oo (x^2)^n = sum_(n=0)^oo x^(2n)`

Multiply by `x` to get:

`x/(1-x^2) = x sum_(n=0)^oo x^(2n) = sum_(n=0)^oo x^(2n+1)`

This is a geometric series with common ratio `x^2`, so it will converge if `abs(x^2) < 1`, which is when `abs(x) < 1`. That is, the radius of convergence is `1`.