I have to find the x values for which it converges.
They are potency series so I just have do use the series without x
a) $\sum_{n=1}^{\infty} \frac{n \sqrt{n}}{3^{n}}(x+3)^{n}$
For a) I don't know if I can use the ratio test for it?
Because it can be 0 for some n?
If I want to use the square root test I am literally stuck at the start.
Our profs told us that we have to be careful and can just use ratio tests if the series will never become 0
b) $\sum_{n=1}^{\infty} \frac{2^{n}}{n \sqrt{n}}(x-2)^{n}$
The ratio test shows that the series converges for $|x+3| <3$ or $-6 <x<0$. [ $\lim \frac {a_{n+1}} {a_n}=3(x+3)$ because $\frac n {n+1} \to 1, \frac {\sqrt n} {\sqrt {n+1}} \to 1, \frac {3^{n+1}} {3^{n}}=3$ and $\lim \frac {(x+3)^{n+1}} {(x+3)^{n}} =x+3$]. The series divers for $x=-6$ and $x=0$ since the general term does not tend to $0$. Recall that $\sum a_n$ cannot converge unless $a_n \to 0$. The radius of convergence is $3$.
b) is similar.