Convergence radius of power series is infinite Which function is given by a power series whose convergence radius is infinite?
$$A. \ \ \ e^{-\frac{1}{x^2}}$$
$$B. \ \ \ \sin{\left(\frac{1}{x}\right)}$$
$$C. \ \ \ \cos{\left(\frac{1}{1+x^2}\right)}$$
$$D.  \ \ \ 1+x+x^3$$ 
$$$$
When we have the power series $\sum_{ n=0}^{\infty} a_n (x−ξ)^n$  the radius of convergence is infinite when $p=0$ , where $p=\lim \sup \sqrt[n]{|a_n |} $. Does this stand? Do I have to find the power series of all these functions? 
 A: You have the right formula for the radius of convergence, but it's rather hard to use in this case. A general fact is that

The radius of convergence of a power series centered at a point is the distance to the closest singularity, be it a pole or an essential singularity.

So figure out which of these functions are entire, and you're done. Polynomials are might nice, though.
A: For D it should be clear, that the convergence radius is $\infty$ (why?).
For A, B and C. U can use the power series extensions of the functions:
$e^x=\sum\limits_{n=0}^{\infty} \frac{x^n}{n!} $
$sin(x)=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$
$cos(x)=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$
(With this way you are also able to classify the pole(s) if u have one)
I give an example:
Given: $f(z)=\frac{e^z}{z}$ The we obtain: $\sum\limits_{n=0}^{\infty} \frac{z^n}{n!}\frac{1}{z}=\sum\limits_{n=0}^{\infty} \frac{z^{n-1}}{n!}=\frac{1}{z}+1+\frac{z}{2!}+\frac{z^2}{3!}+.....$
As we can see we got problems  in 0. In this case 0 is called a pole (pole of order 1), because $\lim_{z \to 0} f(z)=\infty $. So our function can not have a $\infty$ convergence radius.
If you face the situation that there is no addend in the form $\frac{1}{(z-z_0)^n}$, after you have shorten your term, then your function is holomorph on $\mathbb C$ i.e. the convergence radius is $\infty$
Can u now solve your tasks?
