I'm not sure I got all of these problems right. I'd really appreciate any sort of feedback.
For which $x \in \mathbb{R}$ do the following series converge?
Problem 1
For $\sum\limits_{n=1}^{\infty}\frac{3n^7x^n}{2n!} = \sum\limits_{n=1}^{\infty}\frac{3n^7}{2n!}x^n$, perform a ratio test: $$\lim\limits_{n\to \infty}\left|\frac{3(n+1)^7}{2(n+1)!} \cdot\frac{2n!}{3n^7}\right| = \lim\limits_{n\to \infty}\left|\frac{(n+1)^7}{(n+1) \cdot n^7}\right|= \lim\limits_{n\to \infty} \left|\frac{(n+1)^6}{n^7}\right|\\= \lim\limits_{n\to \infty}\left|\frac{n^6+6n^5+15n^4+20n^3+15n^2+6n+1}{n^7}\right| = \lim\limits_{n\to \infty} 0 =0$$
$\implies$ The ratio of convergence is therefore $r = \infty$.
$\implies$ The power series absolutely converges for any $x \in\mathbb{R}$. $$\\$$
Problem 2
For $\sum\limits_{n=1}^{\infty}\frac{(x-1)^n}{n}$, we see that the coefficients $a_n$ are formed by the harmonic series $a_n = \frac{1}{n}$.
Note that $\lim\limits_{n \to \infty} a_n = 0$.
$\implies$ The ratio of convergence is therefore $r = \infty$.
$\implies$ The power series absolutely converges for any $x \in\mathbb{R}$. $$\\$$
Problem 3
For, $\sum\limits_{n=1}^{\infty}\frac{(-4nx)^n}{2n^3} = \sum\limits_{n=1}^{\infty}\frac{(-4n)^n}{2n^3}x^n$ , perform a root test: $$\lim\limits_{n\to \infty}\left|\sqrt[n]{\frac{(-4n)^n}{2n^3}}\right| = \lim\limits_{n\to \infty}\left|\frac{-4n}{\sqrt[n]{2n^3}}\right|= \lim\limits_{n\to \infty}\left|\frac{-4n}{\sqrt[n]{2}\sqrt[n]{n}\sqrt[n]{n}\sqrt[n]{n}}\right| = \infty, \text{ since} \lim\limits_{n\to\infty} \sqrt[n]{n}=1$$. $\implies$ The radius of convergence is $r = \infty$.
$\implies$ The power series will only converge for $x = 0$.
If x=2 in number 2, does the series converge?