Taylor series of $\frac 1 {1+x^2}$ I have to construct the Taylor series of
$$\frac 1 {1+x^2}$$
around $0$ and $1$ and analyze the convergence in both cases. Also (but this is a consequence of the previous series) I have to construct the Taylor series of
$$arctan(x)$$
What I have so far:


*

*I know that the Taylor polynomial around $0$ of $\frac 1 {1+x^2}$ is $$1-x^2+x^4-x^6+....+(-1)^nx^{2n}$$

*If I have the power series of a function f such as $F'=f$, I can construct the power series of $F$ with $F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1}$

 A: 
If I have the power series of a function f such as $F'=f$, I can construct the power series of $F$ with $F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1}$

Have you tried this?  Note that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$.  
You have the power series for $\frac{1}{1+x^2}$ centered at $0$, for which 
$$
a_n = \begin{cases}
(-1)^{n/2} & n \text{ is even}\\
0 & \text{otherwise} 
\end{cases}
$$
In order to find the Taylor expansion of $\frac{1}{x^2 + 1}$ at $1$, note that
$$
\begin{align}
\frac{1}{x^2 + 1} &= 
\frac{1}{1 + (1+(x-1))^2} = 
\frac{1}{1 + 1 + 2(x-1) + (x-1)^2}
\\ &
= \frac{1}{2 + 2(x-1) + (x-1)^2}
= \frac{1}{2} \cdot \frac{1}{1 + \left[(x-1) + \frac{(x-1)^2}{2}\right]}
\\ &
= \frac 12 \left(1 - \left[(x-1) + \frac{(x-1)^2}{2}\right] + \left[(x-1) + \frac{(x-1)^2}{2}\right]^2 - \cdots \right)
\\ &
= \frac 12 - \frac 12 (x-1) + \frac 14(x-1)^2 \\
&\qquad- \frac 18 (x-1)^4 + \frac 18 (x-1)^5 -\frac 1{16} (x-1)^6\\
&\qquad+ \frac 1{32} (x-1)^8 - \frac 1{32} (x-1)^9 + \frac 1{64}(x-1)^{10} - \cdots
\end{align}
$$
