Calculate limit of $\frac{1}{n}\cdot (1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})$ How to prove that
$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$
using only basic limit operations and theorems?

Note: I am not sure what count as a "basic operation" on limits.
Method $1$ (inspired by the classic proof of Integral Test for Series):
Let $S_n=\sum_{k=1}^n \sqrt{k}$. The function $f(x)=\sqrt{x}$ is increasing so, for any integer $k$, we have 
$$\sqrt{k}\leqslant \sqrt{x} \leqslant \sqrt{k+1}$$
for $x\in[k,k+1]$. Integrating the inequality on $[k,k+1]$, we obtain 
$$\sqrt{k}\leqslant \int_{k}^{k+1}\sqrt{x}\; dx\leqslant \sqrt{k+1}$$
Summing from $k=1$ to $n$, we have 
$$S_n \leqslant \int_{1}^{n+1}\sqrt{x} \leqslant S_{n+1}-1$$
The middle integral is $\frac{2}{3}(n+1)^{3/2}-1$, so we obtain that 
$$S_n \geqslant \frac{2}{3}n^{3/2}$$
so $\frac{S_n}{n} \geqslant \frac{2}{3}n^{1/2}$. Since $\lim_{n\to \infty} n^{1/2}=\infty$, we obtain that 
$$\lim_{x\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$

"Method $2$": I believe that the inequality $\frac{S_n}{n} \geqslant \frac{2}{3}n^{1/2}$ can be proved by mathematical induction (but it is way too late in the part of the world I am leaving to try to write it)
