TITLE: Understanding the Taylor expansion of a function
QUESTION [21 upvotes]: Suppose
$$f(x) = \frac{1}{1+x^2}$$
We know this function is defined everywhere and is continuous everywhere and so on...
Using the geometric series, we can write 
$$ \frac{1}{1+x^2} = \sum (-x^2)^n = \sum (-1)^n x^{2n} $$
But, this only converges iff $|-x^2|<1$ iff $|x|<1$. Why does this converge only in $(-1,1)$ when we know it is defined everywhere, though?

REPLY [6 votes]: A more graphical interpretation for this case. Not a proof, more insights. 
Mostly because $f$ is a function that tend to $0$ at $\pm \infty$. This is not a natural behavior for polynomials, and even less so when they have higher degrees. 
However, the function is very smooth far away from $\pm \infty$, that is, around $0$. Here, higher degree polynomials fit the function much better, even more at higher degrees. This is illustrated in the following pîcture, the top one showing that the behavior is very different between $0$ neighborhoods or $+ \infty$, the bottom one only at the vicinity of $0$.

The quality of the approximation, and the radius of convergence, is highly related to the behavior of the coeficients $a_n$ multiplying the terms  $x^n$. Here, the  reason why the border is at $x=1$ is due to the fact you have $|a_n|=1$, which leads to a divergence of the series at $x=\pm 1$. Would the coefficients decrease at a different speed  (for instance $|a_n|\sim \frac{1}{r^n}$), this would modify the radius of convergence to $r$.