TITLE: Maclaurin expansion of arctan: convergence?
QUESTION [6 upvotes]: In my textbook, the Maclaurin series expansion of $\arctan{x}$ is found by integrating a geometric series, that is, by noting that
$\frac{d}{dx}(\arctan(x)) = \frac{1}{x^2+1}$
then rewriting the latter as a geometric series over which one can then integrate. What bothers me is that the geometric series is only convergent when $|x| < 1$, but $\arctan(x)$ is defined for all $x$. This question of convergence is dismissed by the author, but I'm curious as to what's really going on here. Is the series expansion still valid outside the radius of convergence, and if so, why?

REPLY [2 votes]: This is actually an important property of MacLaurin series expansions and highlights the difference between the MacLaurin series expansion of a function and the function itself. When you calculate the Taylor series expansion $a_0+a_1x+ a_2x^2+\cdots$ of a function $f$, what you're actually doing is finding a sequence of polynomials 
$$a_0,\; a_0+a_1x, \;a_0+a_1x + a_2x^2,\ldots$$
approximating $f$. But how well do they approximate $f$? In the case of $f(x)=\arctan x$, they approximate $f$ really well when $|x|\leq 1$, but not very well at all otherwise.