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

Show that: $$ \lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty $$

I've tried the following way. Consider the following sum: $$ \sqrt n + \sqrt{n-1} + \dots + \sqrt{n-\frac{n}{2}} + \dots + \sqrt{2} + 1 $$

Now if we take only $n\over 2$ terms of the sum we obtain that: $$ \sqrt n + \sqrt{n-1} + \dots > {n \over 2} \sqrt{n\over 2} $$ Let: $$ x_n = {1 \over n}(1 + \sqrt{2} + \dots + \sqrt{n}),\ \ n\in \Bbb N $$

Using the above we have that: $$ x_n > {1\over n} {n\over 2}\sqrt{n\over 2} = {1\over 2} \sqrt{n \over 2} $$ Now taking the limit for RHS its obvious that: $$\lim_{n\to\infty}{1\over2}\sqrt{n\over2} = +\infty $$ Which implies: $$ \lim_{n\to \infty}x_n = + \infty $$

Have I done it the right way? Also i would appreciate alternative ways of showing that limit. Thanks!

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.

Just to give an alternative, note first that the sequence is increasing:

$$\begin{align} {1\over n+1}(1+\sqrt2+\cdots+\sqrt n+\sqrt{n+1})-{1\over n}(1+\sqrt2+\cdots\sqrt n) &={\sqrt{n+1}\over n+1}-{1+\sqrt2+\cdots+\sqrt n\over n(n+1)}\\ &\gt{\sqrt{n+1}\over n+1}-{n\sqrt n\over n(n+1)}\\ &={\sqrt{n+1}-\sqrt n\over n(n+1)} \end{align}$$

Now consider

$$\begin{align} {1\over n^2}(1+\sqrt2+\cdots+\sqrt{n^2}) &\gt{1\over n^2}(1+1+1+2+2+2+2+2+3+\cdots+(n-1)+n)\\ &={1\over n^2}(3\cdot1+5\cdot2+7\cdot3+\cdots+(2n-1)(n-1)+n)\\ &\gt{2\over n^2}(1^2+2^2+3^2+\cdots+(n-1)^2)\\ &={2\over n^2}\cdot{(n-1)n(2n-1)\over6}\\ &={(n-1)(2n-1)\over3n}\\ &\to\infty \end{align}$$