Find a limit of a sequence with series Find the limit of given sequence, using only the definition and properties of limits of sequences:
$$
x_{n} = \frac{1 + \sqrt{2} + \sqrt[3]{3} + ... +\sqrt[n]{n}}{n}
$$
I've tried to add and substract n to get:
$$
x_{n} = \frac{(1 - 1) + (\sqrt{2} - 1) + (\sqrt[3]{3} - 1) + ... + (\sqrt[n]{n} - 1) + n}{n}  \le 1 - \frac{C(n - 1)}{n}, 0<C\le1
$$

First off, $\sqrt[k]{k}\geq 1$ for all $k=1,2,...$, so
\begin{align*}
x_{n}\geq\frac{n\cdot 1}{n}=1,\quad n=1,2,...
\end{align*}
On the other hand, since $\sqrt[k]{k}\rightarrow 1$ as $k\rightarrow\infty$, given $\varepsilon>0$, there is an $N\in\mathbb{N}$ such that $\sqrt[k]{k}-1<\varepsilon$, $k\geq N$. Subsequently,
\begin{align*}
x_{n}&=\frac{1+\cdots+\sqrt[N-1]{N-1}}{n}+\frac{\sqrt[N]{N}+\cdots+\sqrt[n]{n}}{n}\\
&\leq\frac{1+\cdots+\sqrt[N-1]{N-1}}{n}+\frac{(n-N+1)}{n}(1+\varepsilon).
\end{align*}
For sufficiently large $n\geq N$, $x_{n}-1<\varepsilon$, so $x_{n}\rightarrow 1$ as $n\rightarrow\infty$.
