Find the Taylor-series for $f$ in $0$

Question

I'm dealing with the following problem.

Let $a\in\mathbb{R}\setminus\{0\}$ and define $f:{\mathbb{R}\setminus\{a\}}$ with

$$f(x)=\frac{1}{a-x}, \hspace{20pt} x\in\mathbb{R}\setminus\{a\}$$

Find the Taylor series for $f$ in $0$.

I can manually calculate for every $n$, but how do I simplify it to a series such as the following which I found using Wolfram?

$$\frac{1}{a-x}=\sum_{n=0}^\infty a^{-1-n}x^n$$

Answer 1

Just write $$ f(x) = \frac{1}{a}\cdot \frac{1}{1-(x/a)} $$ and use the sum of a geometric progression for $\lvert x/a \rvert < 1$.

Answer 2

Hint : Sum of$ (n+1)$ terms of the geometric progression:
$1+y+y^2+y^3+...+y^n = \frac{y^{n+1}-1}{y-1}$
If $|y|\lt 1$ and $n\to \infty $, then we have
$1+y+y^2+...= \frac{1}{1-y}$
In your function $f$, treat $x/a$ as $y$ and obtain Taylor series using the above information.