Power series for $(a+x)^{-1}$ Is it possible to write the following expression in terms of power series?
$$ (a+x)^{-1}=\sum\limits_{k =  - \infty }^\infty  {{b_k}{x^k}} $$
where $0 < a < 1$ and $0 < x < 1$.
 A: It can be turned into geometric series:
$$\frac{1}{a+x} = \frac{1}{a} \cdot \frac{1}{1-\left(-\frac{x}{a}\right)} = \frac{1}{a}\cdot\sum\limits_{k=0}^{\infty}\left(-\frac{x}{a}\right)^k$$
which converges whenever $|x| < |a|$.
A: You can also use the general binomial series (converging for $|x| < 1$):
$$(1+x)^\alpha = \sum_{k=0}^{\infty}\frac{\alpha(\alpha-1) \ldots (\alpha-k+1)}{k!}x^k.$$
Hence,
$$(a+x)^{-1} = a^{-1}(1 + x/a)^{-1}  = a^{-1}\sum_{k=0}^{\infty}\frac{-1(-2) \ldots (-k)}{k!}\left(\frac{x}{a}\right)^k\\ = a^{-1}\sum_{k=0}^{\infty}(-1)^k\left(\frac{x}{a}\right)^k,$$
converging for $|x| < |a|.$
A: $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
$$\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n$$
$$\frac{1}{1+x/a}=\sum_{n=0}^\infty (-1)^n\left(\frac{x}{a}\right)^n$$
$$\frac{1}{a+x}=\frac1a\sum_{n=0}^\infty (-1)^n\left(\frac{x}{a}\right)^n$$
Convergent for $$\left|\frac{x}{a}\right|<1\implies x<a$$
A: Alternatively, there is this geometric series:
$$
\frac{1}{a+x} = \frac{1}{x}\left(\frac{1}{1+\frac{a}{x}}\right)
= \frac{1}{x}\sum_{k=0}^\infty(-1)^k\left(\frac{a}{x}\right)^k
=\sum_{n=-\infty}^{-1}(-1)^{n+1}a^{-n-1}x^n
$$
with convergence for $|x|>a$.
