The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$\lim_{n\rightarrow\infty}n^{\frac1n}=\lim_{n\rightarrow\infty}\exp\left(\frac{\ln n}n\right)=\exp\left(\frac 00\right)=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac1n}1\right)=\exp\left(\frac01\right)=e^0=1$$
And
$$\begin{align}\lim_{n\rightarrow\infty}\left(1+\frac1{n^2}\right)^n&=\lim_{n\rightarrow\infty}\exp\left(n\ln\left(1+\frac1{n^2}\right)\right)\\
&=\lim_{n\rightarrow\infty}\exp\left(\frac{\ln\left(1+\frac1{n^2}\right)}{\frac1n}\right)=\exp\left(\frac00\right)\\
&=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac{-\frac2{n^3}}{1+\frac1{n^2}}}{-\frac1{n^2}}\right)=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac2n}{1+\frac1{n^2}}\right)\\
&=\exp\left(\frac0{1+0}\right)=e^0=1\end{align}$$
Then
$$\lim_{n\rightarrow\infty}\frac{n^{n+\frac1n}}{\left(n+\frac1n\right)^n}=\lim_{n\rightarrow\infty}\frac{n^{\frac1n}}{\left(1+\frac1{n^2}\right)^n}=\frac11=1\ne0$$
So the sum diverges by the divergence test.