For which values of x the following power series converge and for which values of x converge absolutely: $$\sum_{n=0}^{\infty} \frac{n!(2n)!}{(3n)!}x^n$$
I found this question quite strange because I've never seen a question to differ such convergences for the values of x. I only know how to find the values of $x$ in general by the Ratio test.
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|= \lim_{n \to \infty} \left| \frac{(n+1)! \cdot(2n+2)!\cdot x^{n+1}\cdot(3n)!}{(3n+3)! \cdot n! \cdot (2n)!\cdot x^n} \right|=\lim_{n \to \infty} \left| \frac{4n^3+10n^2+8n+2}{27n^3+54n^2+33n+6} \right| |x|=\frac{4}{27}|x|<1 \implies |x|<\frac{27}{4}$$
This is the way I know how to find the values of x so that the series converge absolutely. What about just convergence?
I also faced another problem. If I want to check the end points, I cannot conclude whether power series converge or not like this one:
$$\text{If} \quad x=-\frac{27}{4} \implies \sum_{n=0}^{\infty} (-1)^n \frac{n!(2n)!}{(3n)!} \left( \frac{27}{4} \right)^n$$.
How do I conclude the convergence of such sum?