Continuity and Infinite Sums A well-known (and very useful) theorem in elementary topology states that the sum of two continuous functions is continuous. It can also be proven that any polynomial is continuous.
However, it is also well-known that many functions can be expressed as a MacLaurin Series, which is just an infinite polynomial with coefficients following certain constraints.
This is rather problematic to me. I want to prove that 
$$\cos x$$ is continuous, and I was planning on using this fact to do it. However, the existence of Maclaurin Series for functions like 
$$\frac{1}{1-x}$$
suggests that this is an invalid method of proof.
When can I use Maclaurin Series to prove continuity?
 A: If the radius of convergence $r$ of a series $\sum_{n=0}^\infty a_nx^n$ is greater than $0$, then the function$$\begin{array}{ccc}(-r,r)&\longrightarrow&\mathbb{R}\\x&\mapsto&\displaystyle\sum_{n=0}^\infty a_nx^n\end{array}$$is continuous. In particular, since the radius of convergence of the series$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}$$is $+\infty$, it defines a continuous function. And this function is the cosine function.
A: From complex analysis, if a function of the form
$$f(z)=\sum_{n=0}^\infty a_nz^n$$
converges for all $|z|<R$ but for no $|z|>R$, then there exists a point $|z|=R$ such that $f(z)$ cannot be defined to be continuous at that point (and as far as complex functions go, it is undefined at that point).  Such a point is called a singularity.  Likewise, there cannot exist any singularities within the radius of convergence.  For $f(z)=\frac1{1-z}$, it is fairly obvious that it's Maclaurin expansion has radius of convergence $R=1$, so there exists a point $|z|=1$ such that $f(z)$ has a singularity.  That point happens to be $z=1$.  And $f(z)$ is continuous on $|z|<1$.
For a function such as $f(z)=\frac1{1+z^2}$, it is not obvious from a real standpoint that the Maclaurin expansion has radius of convergence $R=1$, however, from the above, it is easy to see that the function has singularities at $z=\pm i$, hence, $R=\min|\pm i|=1$.
Functions who's Taylor expansions have an infinite radius of convergence are called entire functions, and are continuous everywhere.
