Radius of convergence of $\sum_{n=0}^\infty n!x^{n^2}$
Solution 1:
I think the question is to find the radius of convergence, not to "calculate" the series (I doubt that the sum of the series has a closed-form expression).
Ratio test:
$$ \left|\frac{(n+1)!\; x^{(n+1)^2}}{n!\; x^{n^2}}\right| = (n+1) |x|^{2n+1}$$
If $|x| < 1$ this goes to $0$ as $n \to \infty$, and thus is less than $1$ for sufficiently large $n$, thus the series converges.
If $|x| \ge 1$ it is greater than $1$, and the series diverges.
Thus the radius of convergence is $1$.