Find for which $x, x\in \mathbb{R}$ the series converges. $$\sum_{n=1}^\infty \frac{n+(-3)^n}{\sqrt{n}}(2x)^n$$
I know this is an alternating series and if $x=0$ it's trivial but I don't know how to continue so please can you help me and give some hints please? Thank you.
Update
So if I apply the ratio test I have
$$\lim_{n→∞}\left|\frac{n+1+(-3)^{n+1}}{n+(-3)^{n}}\right||2x| $$ now how can I prove $\lim_{n→∞}\left|\frac{n+1+(-3)^{n+1}}{n+(-3)^{n}}\right|=3 $ ?
It follows from the ratio test that your series converges if $\lvert x\rvert<\frac16$ and diverges if $\lvert x\rvert>\frac16$.
If $x=\frac16$, your series becomse $\displaystyle\sum_{n=1}^\infty\frac{n+(-3)^n}{\sqrt n}\frac1{3^n}$, which converges (it's the sum of two convergent series). And if $x=-\frac16$, your series becomse $\displaystyle\sum_{n=1}^\infty\frac{n+(-3)^n}{\sqrt n}\frac{(-1)^n}{3^n}$, which diverges (it's the sum of a convergent series with a divergent one).