Answer
The interval of convergence for the series is: $(-1, 1]$ and the radius of convergence is $R=1$.
Work Step by Step
Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$
1. For $l \lt 1$, the series is absolutely convergent.
2. For $l \gt 1$, the series is divergent.
3. For $l = 1$, the series is inconclusive.
Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|\\=\lim\limits_{k \to \infty} \dfrac{x^{k+1}}{\sqrt {k+1}} \times \dfrac{\sqrt k}{x^k} \\=\lim\limits_{k \to \infty} \dfrac{\sqrt k}{\sqrt {k+1}}|x|\\= |x|$
So, we can conclude that the given series converges absolutely if $l=|x| \lt 1$ for $-1\lt x \lt 1$. The test is inconclusive if $|x| =1$ for $x=1$ or $x=-1$. It diverges if $l=|x| \gt 1$ .
Therefore, the interval of convergence for the series is: $(-1, 1]$ and the radius of convergence is $R=1$.