Did I calculate these problems on the convergence of power series correctly? I should determine all $x \in \mathbb R$ for which the following power series converge or diverge.
I solved them as follows:
a)
$$\sum \limits_{n=1}^{\infty} \frac{3n^7x^n}{2n!}$$
(Use quotient criterion) $\to$ $$r=\frac{\frac{3n^7}{2n!}}{\frac{3(n+1)^7}{2(n+1)!}} = \frac{n^7}{(n+1)^6} $$ $$\lim\limits_{n\to\infty}\frac{n^7}{(n+1)^6} \rightarrow ∞$$
⇒ The series converges for all $x \in \mathbb R.$
b) $$\sum \limits_{n=1}^{\infty}\frac{(x-1)^n}{n} $$
(use root test) $\to$ $$r=\frac{1}{\lim\limits_{n\to\infty}\sqrt[n]{\frac{1}{n}}} = 1 $$
converges for $0<x<2$ and diverges for $x<0$ and $x>2.$
Boundary points: For $x=0$ the series converges and for $x=2$ the series diverges.
$\to$ The series converges for $x \in [0,2)$
c) $$\sum \limits_{n=1}^{\infty}\frac{(-4nx)^n}{2n^3} = \sum \limits_{n=1}^{\infty}\frac{(-4n)^n}{2n^3} x^n$$
(use quotient criterion) $\to$
$$r=\frac{\frac{(-4n)^n}{2n^3}}{\frac{-4(n+1)^{n+1})}{2(n+1)^3}} = \frac {-(-4)^n *n^{n-3}}{4(n+1)^{n-2}}$$
$$\lim\limits_{n\to\infty} \frac {-(-4)^n *n^{n-3}}{4(n+1)^{n-2}} \rightarrow \text{ no solution possible, the series diverges.}$$
Thank you in advance for your help!!
 A: For the series c): let us consider the series
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( {4n\left| x \right|} \right)^n }}{{2n^3 }}} 
$$
Now we apply the root test for sequences. We have that
$$
\mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{\frac{{\left( {4n\left| x \right|} \right)^n }}{{2n^3 }}}} = \mathop {\lim }\limits_{n \to  + \infty } \frac{{4n\left| x \right|}}{{\left( {2n^3 } \right)^{{\textstyle{1 \over n}}} }} = \left\{ \begin{array}{l}
  + \infty \,\,\,if\,\,\,x \ne 0 \\ 
 0\,\,\,\,\,\,\,if\,\,x = 0 \\ 
 \end{array} \right.
$$
we have that, for $x \neq 0$,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{{\left( {4n\left| x \right|} \right)^n }}{{2n^3 }} =  + \infty 
$$
Thus, for $x \neq 0$ it can't be
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{{\left( { - 4nx} \right)^n }}{{2n^3 }} = 0
$$
which is the necessary condition for the convergence of the given series. Therefore, the series diverges for every $x \neq 0$. Of course, it converges trivially if $x=0$
By the way you can't apply the quotient criterion if you don't have a series with positive terms.
