How to calculate the radius and interval of convergence of the following series: $$\sum 2^{n^2}x^{n!}$$
The formula for the radius is: $$R = \frac{1}{\limsup_{n\to\infty} \sqrt[n]{|a_n|}}$$ or (that is if the limit exists): $$R = \lim_{n\to\infty}\bigg|\frac{a_n}{a_{n+1}}\bigg|$$
but as for now I always had series of the following form:
$$\sum a_n(x-c)^n$$
but now I have factorials $x^{n!}$ and $n^2$ in $a_n$, how do I approach this? The result itself is not very important to me but I want to understand how to approach problems where I have $x^{f(n)}$ instead of $x^n$.
Any hints, suggestions, etc. are welcome!
$$a_k = \begin{cases} 2^{n^2} &, k = n!\\ 0 &, \text{otherwise}\end{cases}$$
You have a lot of zero coefficients in the series. These don't matter for the radius of convergence, so you have
$$R = \frac{1}{\limsup \sqrt[n!]{2^{n^2}}} = 1.$$
If you have a series $$\sum_{n=0}^\infty c_n z^{a(n)}$$ with a strictly increasing $a\colon \mathbb{N}\to\mathbb{N}$, the radius of convergence is determined as
$$\frac{1}{R} = \limsup_{n\to\infty} \lvert c_n\rvert^{1/a(n)},$$
since only the nonzero coefficients matter.