TITLE: Is there a deeper pattern in the radius of convergence?
QUESTION [8 upvotes]: I'm curious why the Taylor expansion of $\sin(x)$ and $\cos(x)$, for example, converge everywhere on the domain. But the expansion of $\frac{1}{1-x}$ converges for $|x| < 1$. I can compute it trivially using the ratio test and understand that the terms must decrease geometrically with the expansion for $\frac{1}{1-x}$ to convergence. But my question is more geared towards why this is the case: why some functions have power series' that converge everywhere and some that converge only on a certain interval? In other words, besides doing the ratio test, I don't see any sort of pattern or reasoning for why this is the case. Is there some higher level explanation someone can give to a person like myself who hasn't taken real or complex analysis, if the answer lies in those fields? I'm already a bit familiar with the fact that it is called the radius of convergence because it converges on a disc in the complex plane, but nothing further. Thank you.

REPLY [14 votes]: The answer is yes, there is some simple explanation of why this happens. To understand this better, you need to take a class in Complex Analysis.
Fact 1 A real function is equal to a power series $\sum_{n=0}^\infty a_n(x-z)^n$ on $(a-R, a+R)$ if and only if, there exists a complex function $g$, which is differentiable on $|z-a|<R$ such that $f=g$ on $(a-R,a+R)$.
Fact 2 A complex function is equal with its Taylor series on $|z-a|<R$ if and only if it is differentiable on $|z-a|<R$.
It follows from Fact 2, that the radius of convergence for an analytic function (real or complex) is the distance to the closest point in the complex plain where the function is not differentiable.
Since $\sin(x), \cos(x)$ can be extended to complex functions which are analytic everywhere, the radius of convergence is $\infty$. 
For $\frac{1}{1-x}$, or even more interesting $\frac{1}{1+x^2}$, the radius of convergence at $a=0$ is simply the distance to $1$ respectively $\pm i$, which is $1$.