Find real numbers $x$ such that $\sum_{n=0}^{\infty}n!x^{n!}$ converges and absolutely converges. 
Find real numbers $x$ such that the series converges and absolutely converges.
$$\sum_{n=0}^{\infty}n!x^{n!}$$

How do I do this? Can somebody show how it is done when treating $\{n!x^{n!}\}$ as a subsequence of $\{nx^{n}\}$?
 A: One might as well calculate the radius of convergence of the series. Set $a_m=0$ if $m\notin\{n!:n\in\mathbb{Z}_+\}$ and $a_m=m$ otherwise. Then the radius of convergence of the corresponding power series is
$$R=\frac{1}{\limsup_m\sqrt[m]{|a_m|}}=1$$
Thus, the power series converges for all $z$ in the unit disc centered at $0$. Diverges fo all $z$ with $|z|>1$. If $|z|=1$, as $\limsup_m|a_mz^m|=\limsup_m|a_m|=\infty$, the series diverges.

A: For $\vert x \vert \ge 1$ the series $\sum a_n(x)$ with
$$a_n(x) =n!x^{n!}$$ diverges as its term doesn’t converge to zero.
Otherwise
$$\left\vert \frac{a_{n+1}(x)}{a_n(x)}\right\vert =(n+1)\vert x \vert^{n \cdot n!} \le (n+1)\vert x \vert^n$$ As the right end side is less than $1/2$ for $n$ large enough, the series $\sum a_n(x)$ converges absolutely for $\vert x \vert \lt 1$ by comparison with a geometric series.
A: The ratio of terms is $n x^n,$ which is smaller than one (eventually) in absolute value when $|x|<1,$ and only then.
A: If $|x|\ge 1$ then the term does not go to $0$.  DIVERGE
For $|x| < 1$ try ratio test:
$$
\frac{(n+1)! |x|^{(n+1)!}}{n!|x|^{n!}} = (n+1)|x|^{(n+1)!-n!} = (n+1)|x|^{nn!} \le (n+1) |x|^n \to 0
$$
CONVERGE
