Power Series for $(1+x)^{-1}$

We will form the power series centered about $x=0$ using Taylor's formula

$$f(x)=\displaystyle \sum^{\infty}_{i=0}\displaystyle \frac{f^{(i)}(0)x^i}{i!}$$

First we list the first few derivatives of $(1+x)^{-1}$.

$$f(x)=\displaystyle \frac{1}{1+x} \longrightarrow f(0)=1$$

$$f'(x)=-\displaystyle \frac{1}{(1+x)^2} \longrightarrow f'(0)=-1$$

$$f''(x)=\displaystyle \frac{2}{(1+x)^3} \longrightarrow f''(0)=2$$

$$f'''(x)=-\displaystyle \frac{2*3}{(1+x)^4} \longrightarrow f'''(0)=3!$$

$$\vdots$$

$$f^{(n)}(x)=(-1)^n\displaystyle \frac{n!}{(1+x)^{n+1}} \longrightarrow f^{(n)}(0)=(-1)^nn!$$

Seeing the pattern and using Taylor's formula we get

$$(1+x)^{-1}=\displaystyle \sum^{\infty}_{i=0}(-1)^n\displaystyle \frac{n!x^n}{n!}= \displaystyle \sum^{\infty}_{i=0}(-1)^nx^n$$

We formed this power series because we are interested in the power series for $(1+x^2)^{-1}=\displaystyle \frac{1}{1+x^2}$. To find this power series we simply need to replace $x$ with $x^2$ in the first series.

$$\displaystyle \frac{1}{1+x}=\displaystyle \sum^{\infty}_{i=0}(-1)... ...^n=\displaystyle \sum^{\infty}_{i=0}(-1)^nx^{2n}= \displaystyle \frac{1}{1+x^2}$$

What about the radius of convergence? We will use the ratio test.

$$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^{2n+2}}{x^{2n}}\right\vert$$

$$= \displaystyle \lim_{n\to\infty}x^2$$

$$<1 \; \;-1<x<1$$

Clearly it does not converge at either endpoint $x=\pm1$.

We could have also realized that this was an alternating series whose terms approach zero only when $-1<x<1$.