calculus - Radius of convergence of $sum_{n=1}^infty(n!)^2x^{n^2}$ and $sum_{n=1}^infty frac {x^{n^2}}{n!}$

This is a question from an exam I recently failed. 😢

What is the radius of convergence of the following power series? $$(a) \sum_{n=1}^\infty(n!)^2x^{n^2}$$ and $$(b) \sum_{n=1}^\infty \frac {x^{n^2}}{n!}$$

Edit: Here's my attempt at the first one, if someone could tell me if it's any good...

$\sum_{n=1}^\infty(n!)^2x^{n^2}$=$\sum_{n=1}^\infty a_nx^n$ where $a_n= 0$ for any $n\notin \{k\in N| \exists t\in N: k=t^2\}$ and $a_n = (n!)^2$ else. So the radius of convergence would be the inverse of $\lim_{n\rightarrow \infty}{(n!)^{2/n}}=\lim e^{2/n\cdot log(n!) }$. The exponent with log of factorial becomes a series, $\sum_{n=1}^{\infty} \frac{logn}{n}$ which diverges by comparison test with $\frac{1}{n}$, so the radius of convergence would be equal to $0$.

Second edit: This is wrong. Correct answer below.

Answer

Let's consider the first series $\sum_{n=1}^{\infty} (n!)^2 x^{n^2}$. The easiest way to find the radius of convergence is to forget this is a power series and treat $x$ as a constant. Let's assume $x > 0$ so that the terms of the series are positive and we can use any test we wish for the convergence/divergence of a non-negative series. Alternatively, replace $x$ with $|x|$ and check for absolute convergence instead. Using the ratio test, we get the expression
$$\frac{(n+1)!^2 x^{(n+1)^2}}{(n!)^2 x^{n^2}} = (n+1)^2 x^{2n + 1}.$$
This expression converges if $0 < x < 1$ to $0$ while it diverges if $x \geq 1$. This implies that the series converges if $0 < x < 1$ and diverges if $x \geq 1$. But this is a power series so the only possible value for the radius of convergence is $R = 1$ (because it should converge when $|x| < R$ and diverge when $|x| > R$).