Radius of convergence of $\sum_{k=1}^{\infty} {{k^{\sqrt k}}x^{k}}$

Following your approach by ratio test we have that

$$\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt k}}}=\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt {k+1}}}}\frac{{k^{\sqrt {k+1}}}}{{k^{\sqrt k}}}=\left(1+\frac1k\right)^{\sqrt {k+1}}k^{(\sqrt {k+1}-\sqrt k)}\to 1$$

indeed

$$\left(1+\frac1k\right)^{\sqrt {k+1}}=\left[\left(1+\frac1k\right)^k\right]^{\frac{\sqrt {k+1}}{k}}\to e^0=1$$

and

$$k^{(\sqrt {k+1}-\sqrt k)}=e^{(\sqrt {k+1}-\sqrt k)\log k}=e^{\frac{\log k}{\sqrt {k+1}+\sqrt k}}\to e^0=1$$

From the root test we have

$$\lim_{k\to \infty}\sqrt[k]{\left|k^{\sqrt k}x^k\right|}=|x|\lim_{k\to\infty}k^{1/\sqrt{k}}=|x|$$

Can you finish?