Radius of convergence of $\sum_{n = 0}^{\infty}{2^{-n^{2}}n!}x^n$? Given a power series $\sum_{n = 0}^{\infty}{2^{-n^{2}}n!}x^n$, how can I calculate the radius of convergence?
I know that the radius of convergence is $\frac{1}{{\limsup{|c_n|}^\frac{1}{n}}}$ for the coefficient sequence $c_n$, but I don't understand how to find that in this case.  $c_n = {2^{-n^{2}}n!}$, but how can I find the $\limsup$ of this?  When I try the Ratio Test, I get $\frac{x(n+1)}{2^{2n + 1}}$, and I can conclude that as $n$ goes to infinity, this goes to $0$. Does tat mean that the radius of convergence is true for all radii greater than $0$?
 A: Using just the comparison test: Note that
$$\tag 1 |n!x^n/2^{n^2}| \le n^n|x|^n/2^{n^2} = (n|x|/2^n)^n.$$
Now for any fixed $x,n|x|/2^n \to 0.$ Thus for large $n, n|x|/2^n < 1/2.$ So for such $n$ the right side of $(1)$ is less than $(1/2)^n.$ Since $\sum_n (1/2)^n < \infty,$ the power series converges absolutely for this $x$ by the comparison test.  Since $x$ was any real number, the power series converges for all $x,$ hence its radius of convergence is $\infty.$ 
A: By Cauchy-Hadamard Theorem, the radius of convergence $r$ of a series $\sum_{j=0}^\infty a_j z^j$ satisfy
$$ \frac{1}{r} = \limsup_{j \to \infty} \sqrt[j]{|a_j|}
$$
with the understanding of $0^{-1} = \infty, \infty^{-1} = 0$. $r = \infty$ means that $\sum_{j=0}^\infty a_j z^j$ converges for all $z \in \mathbb{C}$. To find the $\limsup$, notice that whenever the limit exists,
$$ \liminf_{j \to \infty} |a_j| = \lim_{j \to \infty} |a_j| = \limsup_{j \to \infty} |a_j|
$$
As for the ratio test, it can be shown that
$$ \liminf_{j \to \infty} \frac{|a_{j+1}|}{|a_j|} \leq \liminf_{j \to \infty} \sqrt[j]{|a_j|} \leq \limsup_{j \to \infty} \sqrt[j]{|a_j|} \leq \limsup_{j \to \infty} \frac{|a_{j+1}|}{|a_j|}
$$
hold for arbitrary sequences $(a_j)_{j=0}^\infty$. Since the ratio test sandwiches the root test, your reasoning is valid.
