The Radius of Convergence of $f(x) = \sum_{n=0}^{\infty} {a_n(x − a)^n}$ Let $f(x) = \frac{1}{1+x^2}$. Consider its Taylor expansion about a point $a ∈ \mathbb{R}$, given by $$f(x) = \sum_{n=0}^{\infty} {a_n(x − a)^n}$$ What is the radius of convergence of this series?    

To solve this I need to find $f^{(n)}(a)$ but I can't find any general formula to calculate $f^{(n)}(a)$. Can I get some help?
 A: HINT. Recall the Taylor Series for $1/(1+x)$ about $x=0$ is given by
$$
\sum_{i=0}^\infty (-1)^i x^i
$$
for $-1<x<1$.
A: HINT 
In order to compute the successive derivatives, I suggest you write $y=x^2$. Then
 $f =\frac{1}{1 + y}$,
 $\frac{df}{dx} = \frac{df}{dy} \frac{dy}{dx}$
and repeat this process. The successive derivatives of $f$ with respect to $y$ are very simple to set.  
Then, the expansion will write
$\sum_{n=0}^{\infty} (-1)^n A(n) \frac{(x - a)^n}{(1 + a^2)^n}$
$A(n)$ being a polynomial of $a$ of degree $n$  
I am sure you can continue from here.
A: 
Hint: Decompose the fraction into simple elements.

Let $x=a+u$, then $1+x^2=(x-i)(x+i)$ hence
$$
2if(x)=\frac1{u+a-i}-\frac1{u+a+i}=\frac1{a-i}\frac1{1+\frac{u}{a-i}}-\frac1{a+i}\frac1{1+\frac{u}{a+i}}.
$$
Expanding both $1/(1+v)$ rational functions as power series in $v$ when $|v|\lt1$, one gets
$$
2if(x)=\sum_{n\geqslant0}(-1)^n\left(\frac1{(a-i)^{n+1}}-\frac1{(a+i)^{n+1}}\right)\,u^n,
$$
that is,
$$
f(x)=\sum_{n\geqslant0}(-1)^n\frac{(a+i)^{n+1}-(a-i)^{n+1}}{(2i)(a^2+1)^{n+1}}\,u^n.
$$
To further identify the coefficients of this expansion, note that
$$
(a\pm i)^{n+1}=\sum_k{n+1\choose k}(\pm1)^ki^ka^{n+1-k},
$$
hence
$$
(a+i)^{n+1}-(a-i)^{n+1}=2i\sum_k{n+1\choose 2k+1}(-1)^ka^{n-2k},
$$
and finally,
$$
f(x)=\sum_{n\geqslant0}A_n(a)(x-a)^n,
$$
where
$$
A_n(a)=\frac{(-1)^n}{(a^2+1)^{n+1}}\sum_{0\leqslant2k\leqslant n}{n+1\choose 2k+1}(-1)^ka^{n-2k}.
$$ 
The expansion is valid when $v=u/(a\pm i)$ is such that $|v|\lt1$, that is, for $|x-a|\lt\sqrt{a^2+1}$.
As a sanity check of the final result, note that $A_{2n}(0)=(-1)^n$ and $A_{2n+1}(0)=0$ for every $n$.
