# Finding radius of convergence for $\frac{x^n}{n\sqrt{n}12^n}$

I need help finding the radius of convergence for $$\frac{x^n}{n\sqrt{n}12^n}$$

I tried both the root and ratio tests but neither one is getting me anywhere.

Trying the root test I reduced to $$\lim \limits_{n \to \infty}\left(\frac{x}{n^{1/2}n^{1/4}12}\right)$$ which is either incorrect or not helpful.

I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.

• Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$ – stochasticboy321 Dec 6 '18 at 1:42

$$\lim_{n \to \infty}\frac{x^{n+1}}{(n+1)\sqrt{n+1}\cdot 12^{n+1}}\frac{n\sqrt{n} \cdot 12^n}{x^n}= \lim_{n \to \infty}\frac{x}{12}\lim_{n \to \infty}\frac{n}{n+1}\lim_{n \to \infty}\sqrt{\frac{n}{n+1}}$$
• Second one goes to $1$, third one also $1$? – SolidSnackDrive Dec 6 '18 at 1:50
• yes, $\frac{n}{n+1}= \frac{1}{1+\frac1n}$. – Siong Thye Goh Dec 6 '18 at 1:51
The ratio test should work. The ratio of the $$1/(n \sqrt(n))$$ part goes to $$1$$ and the rest gives $$|x/12|$$.