Proof verification for $\lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty$

That's seems fine, the more straightforward alternative way is by Stolz-Cesaro, that is

$$\frac{1+\sqrt2+\dots + \sqrt{n+1}-(1+\sqrt2+\dots + \sqrt{n})}{n+1-n}=\sqrt{n+1}$$

As another one alternative, we can use AM-GM

$$\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) \ge \sqrt[2n]{n!}$$

$\bigl(1+\sqrt2+\dots + \sqrt{n}\bigr) $ is an upper Riemann sum, with the subdivision $\{0,1,2,\dots ,n\}$, for the integral $\;\displaystyle \int_0^n\sqrt x\,\mathrm d x=\frac23n^{3/2}$,so $$\frac1n\bigl(1+\sqrt2+\dots + \sqrt{n}\bigr)\ge\frac1n\int_0^n\sqrt x\,\mathrm d x=\frac23\sqrt n,$$ which tends to $+\infty$.

In style to @Bernard's answer, but using this general (and very useful) trick $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$ where $f(x)=\sqrt{x}$, we have $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{k=1}^n \sqrt{\frac{k}{n}}= \int\limits_{0}^{1} \sqrt{x}dx=\frac{2}{3} \tag{1}$$ Now $$\frac{1}{n}\sum\limits_{k=1}^n \sqrt{k}=\sqrt{n}\left(\frac{1}{n}\sum\limits_{k=1}^n \sqrt{\frac{k}{n}}\right)\overset{(1)}{>}\sqrt{n}\left(\frac{2}{3}-\varepsilon \right) \tag{2}$$ from some $n_0$ onwards. And the result follows.