Finding power series representation of the fucntion $f(x) = \frac{1+x^2}{1-x^2}$.

Question

So I want to find the power series representation of the function $f(x) = \frac{1+x^2}{1-x^2}. $ So how do I solve this? I can start by reducing the function to the known form of geometric series but I have a summation at the numerator. How do I overcome that?

$\frac{1+x^2}{1-x^2} = \frac{(1+x)^2 -2x}{1-x^2} = \frac{1+x}{1-x} - \frac{2x }{1-x^2}$, now second part I can solve, it would be $2x \sum_{n =0}^{\infty} x^{2n}.$ How do I go about the first part? Please give some hint. Thank You.

Answer 1

I would go straight using with $X=x^2$

$${1\over 1-X}=\sum_{n=0}^\infty X^n$$

This leads to

$${1+x^2\over 1-x^2}=\left(1+x^2\right)\sum_{n=0}^\infty x^{2n}$$

Now rearranging the sum

$${1+x^2\over 1-x^2}=1+2\sum_{n=1}^\infty x^{2n}$$

Answer 2

Simpler:

Substitute $u=x^2$ in $$\frac{1+u}{1-u}=(1+u)\sum_{k=0}^\infty u^k=\sum_{k=0}^\infty u^k+\sum_{k=1}^\infty u^k.$$