Solution: We know that if $\displaystyle \lim_{n \to \infty} a_n = a$, then \[ \lim_{n \to \infty} \frac{a_1 + a_2 + \dots + a_n}{n} = \lim_{n \to \infty} a_n. \] Therefore, \[ \lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n} = \lim_{n \to \infty} \sqrt[n]{n}. \] Let us find the limit of $\sqrt[n]{n}$ as $n$ approaches to infinity. For $n \gt 1$, since $\sqrt[n]{n} \gt 1$, \[ \sqrt[n]{n} = 1 + h_n, \] where $h_n$ is a positive sequence of real numbers. Thus, \begin{align*} & n = \left( 1 + h_n \right) ^n \geq \frac{n(n-1)}{2}h_n^2 \\ \implies & n \geq \frac{n(n-1)}{2}h_n^2 \geq 0 \\ \implies & \frac{2}{n-1} \geq h_n^2 \geq 0. \end{align*} So by Sandwich's theorem, we conclude that $h_n^2 \rightarrow 0$ and hence $h_n \rightarrow 0$. Therefore, \[ \lim_{n \to \infty} \sqrt[n]{n} = 1. \] Hence, \[ \lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n} = 1. \]