Why Taylor series of $\arctan(x)$ is $\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$. Why Taylor series of $\arctan(x)$ is $\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$. 
I understand that for $x \in [-1,1], \arctan(x)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$. A proof of this fact uses Abel's theorem and an observation that $$\arctan(x)=\int_{0}^{x} \frac{1}{1+t^2}dt=\int_{0}^{x} \sum_{n=0}^{\infty}(-t^2)^ndt= \sum_{n=0}^{\infty} (-1)^n\int_{0}^{x} (t^{2n}) dt$$ 
But how do I see that $$\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1} =\sum_{n=0}^{\infty} \frac{\arctan^{(n)}(a)}{n!}(x-a)^n$$ for some $a$.
Edit
I think I got it. We know that for $x \in [-1,1], \arctan(x)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$, in other words on $[-1,1]$ $\arctan$ function is the power series. We know that a power series $f=\sum_{n=0}^{\infty}c_n(x-a)^n$ which converges on $(a-R,a+R)$ with $R>0$ is infinitely differentiable on $(a-R,a+R)$ and $f^{(n)}(a)=c_n n!$. So $\arctan^{(2n+1)}(0)=(-1)^n \frac{1}{2n+1} (2n+1)!$ and $\arctan^{(2n)}(0)=0$. Now if we write down the Taylor series of $\arctan$ about $0$ we get exactly $\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$.
Thanks to all who helped me to get it with their comments.
 A: As so many people, you seem confused into believing one should use convergent series to understand a property of formal power series. Not so; formal power series are actually quite a bit easier to understand all by themselves. And a Taylor series is a formal power series: it just collects the information about repeated derivatives at a single point (here $0$), in a particular format. (It happens that $\arctan$ is an analytic function, so its values sufficiently close to the point$~0$ are given by evaluating its Taylor series, which converges there; this information is however quite independent from what you are asking about.)
You know that the derivative of $\arctan$ is $x\mapsto\frac1{1+x^2}$. As a rational function defined for $x=0$, the latter is indefinitely differentiable at $x=0$, and therefore so is $\arctan$; then both functions possess a Taylor series at $0$. The Taylor series of $x\mapsto\frac1{1+x^2}$, when multiplied by (the formal power series) $1+x^2$ must give the constant series $1$ (because the Taylor series of a product is the product of the Taylor series), and there is only one formal power series with that property, the series $\sum_{n\in\Bbb N}(-x^2)^n=\sum_{i\geq0}(-1)^nx^{2n}$; this must be the Taylor series of $x\mapsto\frac1{1+x^2}$ at$~0$. It is also the term-by-term derivative of the Taylor series of $\arctan$, which then must be $\sum_{i\geq0}\frac{(-1)^n}{2n+1}x^{2n+1}$.
As you can see, almost no calculus is used, beyond the statement about the derivative of $\arctan$.
A: If you have done some complex analysis arctan is sometimes defined as 
$$\arctan(z) =\frac{\log(z-i)-\log(z+i)}{2i}$$
Then you can use the Taylor expansion of log to prove it. Complex analysis also provides an interesting explanation for why the convergence is on $z \in [-1,1]$ : An analytic function has power series representation with convergence on any disc with radius smaller than distance to the pole closest to the center of the expansion.
