Can anybody help to find radius of convergence for series? $$\sum_{n=1}^{\infty}\frac{(-3)^{n}+2^{n}}{\sqrt[3]{n+1}}(x-2)^n$$
As I used Cauchy ratio test I get such expressions:
$$\lim_{x\to\infty}\sqrt[n]{\frac{(-3)^{n}+2^{n}}{\sqrt[3]{n+1}}(x-2)^n}=\lim_{x\to\infty}\frac{\sqrt[n]{(-3)^{n}+2^{n}}}{(n+1)^{1/3n}}(x-2)$$
As $$\lim_{x\to\infty}(n+1)^{1/3n}=1$$
I have to work with nominator: $$\lim_{x\to\infty}{\sqrt[n]{(-3)^{n}+2^{n}}}(x-2)$$
But here I don't know what to do it's like $$\sqrt[\infty]{(\infty -\infty)}(x-2)$$
Help me, please.
 A: Cauchy-Hadamard Formula:
$$\sqrt[n]{\left|\frac{(-3)^n+2^n}{\sqrt[3]{n+1}}\right|}=\frac1{(n+1)^{1/3n}}\cdot 3\cdot\sqrt[n]{\left|1+\left(-\frac23\right)^n\right|}\xrightarrow[n\to\infty]{}\frac11\cdot3\cdot1=3$$
So the radius is $\;R=\cfrac13\;$
Another way: the ratio test
$$\left|\frac{(x-2)^{n+1}\left[(-3)^{n+1}+2^{n+1}\right]}{\sqrt[3]{n+2}}\right|\left|\frac{\sqrt[3]{n+1}}{(x-2)^n\left[(-3)^n+2^n\right]}\right|=$$
$$=\sqrt[3]\frac{n+1}{n+2}\cdot3|x-2|\cdot\frac{1+\left(-\frac23\right)^{n+1}}{1+\left(-\frac23\right)^n}\xrightarrow[n\to\infty]{}3|x-2|$$
so we must have
$$|x-2|<\frac13$$
A: hint
compute the two limits
$$L_1=\lim_{\infty} |a_{2n}|^{\frac{1}{2n}} $$
and
$$L_2=\lim_{\infty}|a_{2n+1}|^{\frac {1}{2n+1}} $$
then $$R=\frac {1}{\max (L_1,L_2)} . $$
A: First mistake: in the Root Test for power series, you must use the absolute values of the terms, not just the terms themselves. So what you set up here must look as
$$\lim_{n\to\infty} \sqrt[n]{\left|\frac{(-3)^n+2^n}{\sqrt[3]{n+1}}(x-2)^n\right|}=\lim_{n\to\infty} \frac{\sqrt[n]{\left|(-3)^n+2^n\right|}}{(n+1)^{1/3n}}|x-2|.$$
Also note that the limit is taken as $n\to\infty$, not as $x\to\infty$.
Then yes, this boils down to finding the limit $\lim\limits_{n\to\infty} \sqrt[n]{\left|(-3)^n+2^n\right|}$ (which will be multiplied by $|x-2|$). As suggested in comments, you can factor our $3^n$ to find
$$\lim\limits_{n\to\infty} \sqrt[n]{\left|(-3)^n+2^n\right|}=\lim\limits_{n\to\infty} \sqrt[n]{3^n\left|(-1)^n+(2/3)^n\right|}=\lim\limits_{n\to\infty} 3\cdot\sqrt[n]{\left|(-1)^n+(2/3)^n\right|}=3,$$
because $(2/3)^n\to0$ as $n\to\infty$. So you'll get that
$$\lim_{n\to\infty} \sqrt[n]{\left|\frac{(-3)^n+2^n}{\sqrt[3]{n+1}}(x-2)^n\right|}=\lim_{n\to\infty} \frac{\sqrt[n]{\left|(-3)^n+2^n\right|}}{(n+1)^{1/3n}}|x-2|=3|x-2|.$$
