Problem: The limit
\[
\lim_{n \to \infty} \frac{1}{n}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n} \right)
\]
- is equal to $0$
- is equal to $1$
- is equal to $2$
- does not exist
Solution: According to the Cauchy's limit theorem,
(Cauchy's Limit Theorem) If $\left( a_n \right) $ is a sequence such that $a_n \to a$, then
\[
\lim_{n \to \infty} \frac{a_1 + a_2 + \dots + a_n}{n} = a.
\]
Here $a_n = \sqrt[n]{n} $. So
\[
\lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n} = \lim_{n \to \infty} \sqrt[n]{n}.
\]
We will prove that $\sqrt[n]{n} \rightarrow 1$. If $n>1$, then $\sqrt[n]{n} > 1$. Thus, for each $n > 1$, we can write
\[
\sqrt[n]{n} = 1 + h_n,
\]
where $h_n$ is a positive number. Thus,
\begin{align*}
\sqrt[n]{n} = 1 + h_n & \implies n = \left( 1 + h_n \right) ^n \\
& \implies n = 1 + n h_n + \frac{n(n-1)}{2}h_n^2 + \dots + h_n^n \\
& \geq \frac{n(n-1)}{2}h_n^2.
\end{align*}
So we obtained
\begin{align*}
0 \leq \frac{n(n-1)}{2} h_n^2 \leq n & \implies 0 \leq h_n^2 < \frac{2}{n-1}.
\end{align*}
From the Sandwich's theorem,
\[
\lim_{n \to \infty} h_n^2 = 0 \implies \lim_{n \to \infty} h_n = 0.
\]
Therefore,
\[
\lim_{n \to \infty} \sqrt[n]{n} = 1
\]
and hence,
\[
\lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n} = 1.
\]