Convergence radius and interval

I need to find radius of convergence and interval of convergence of $\displaystyle \sum_{n=1}^{\inf}\frac{x^n}{\sqrt{n}}$

In d'Alembert feature I find radius like this $\displaystyle \lim_{n->\inf}\frac{x^{n+1}\sqrt{n}}{\sqrt{n+z}x^n}=|x|< 1$

$\displaystyle |x|<1$

$\displaystyle -1<x<1$

Radius is 1

If I try in Cauchy feature, I get same:

$\displaystyle \lim_{n->\inf}\sqrt{|\frac{x^n}{\sqrt{n}}|}<1$ first square $\displaystyle n$ root

$\displaystyle \lim_{n->\inf}|\frac{x}{\sqrt_{n+2}{n}}}|<1$ square $\displaystyle n+2$ root

$\displaystyle |x|<1$

$\displaystyle -1<x<1$

Radius is 1

Interval of convergence:

I take -1 and 1

$\displaystyle \sum_{n=1}^{\inf}\frac{x^2 * 1^n}{\sqrt{n}}$ and $\displaystyle \sum_{n=1}^{\inf}\frac{x^2 * (-1)^n}{\sqrt{n}}$

I do check if diverge/converge:

$\displaystyle \lim_{n->\inf}\frac{x^n}{\sqrt{n}}=\inf$ it diverge

$\displaystyle \lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0$ here **!=** means **'not even'**

Interval is $\displaystyle x\E(-1;1)$

Questions:

- Did I do task correct?
- $\displaystyle \lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0$ it diverge or converge?
- How interval of convergence change because of divergence and convergence?