Separately check the \({\rm{n}}\)component.
\(\begin{array}{c}{\rm{y = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{n}}^{{\rm{2/n}}}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\rm{ln}}{{\rm{n}}^{{\rm{2/n}}}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{2}}}{{\rm{n}}}{\rm{lnn}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{2}}\frac{{\rm{1}}}{{\rm{n}}}}}{{\rm{1}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{2}}}{{\rm{n}}}\\{\rm{ = 0y}}\\{\rm{ = }}{{\rm{e}}^{\rm{0}}}\\{\rm{ = 1}}\end{array}\)
Substitute in equation \({\rm{(1)}}\)
\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{25}}}}{{{{\rm{n}}^{{\rm{2/n}}}}{\rm{9}}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{25}}}}{{{\rm{(1)9}}}}\\{\rm{ = }}\frac{{{\rm{25}}}}{{\rm{9}}}\end{array}\)
As a result, the Root Test shows that the provided series is divergent.