For $a \in \mathbb{R}\setminus\{0\}$ we have function $f$:$\mathbb{R}\setminus\{a\}\rightarrow \mathbb{R}$:
$$f(x)=\frac{1}{a-x}$$ for $x\in \mathbb{R}\setminus\{a\}$.
Then I have to find the Taylor serie in $0$. I think it's : $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{a^{n+1}}$.
I also have to find the radius of convergence. I think it's: so $r=a$.
But how can I check convergence in the edge? I have tried for quotient test. I got for x=a 1/a and for x=-1 -1/a. What will the conclusion be then? We have convergent on the edge for some $a$?
$\frac 1 a$ is not an end point of the interval of convergence.
The series converges for $-|a|<|x|<|a|$. The boundary points of the interval of convergence are $\pm a$. Clearly the series diverges at these points. [Let $x =\pm a$. Note that $|\frac {x^{n}} {a^{n+1}}|=|\frac 1 a|$ so $\frac {x^{n}} {a^{n+1}}$ does not tend to $0$ as $n \to \infty$. Hence the series is divergent when $x =\pm a$].