Find Taylor serie and radius of convergence

Question

For a$\in$R/{0} we have function $f$:$R$/{a}$->$R: $$f(x)=\frac{1}{a-x}$$ for x$\in$$R$/{a}. Then I have to find the Taylor serie in 0. I think it's maybe: $\displaystyle\sum_{n=0}^{\infty}(\frac{n!}{a^{n+1}}/n!) x^n$. It's that correct? How can I show it?

I also have to find the radius of convergence. I think it's: $r^{-1}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{ \frac{(n+1)!}{a^{n+2}}/(n+1)! }{ \frac{n!}{a^{n+1}}/n!}=1/a$ so r=a? Is that correct? And how can I check convergence in the edge?

Answer 1

For $a \ne 0$ we have

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

for $|x|<|a|$ .

Geometric series !

The enpoints:

is $\sum_{n=0}^{\infty}\frac{x^n}{a^n}$ convergent for $x=a ?$

is $\sum_{n=0}^{\infty}\frac{x^n}{a^n}$ convergent for $x=-a ?$