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!
 A: $\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 $+\infty$.
A: 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.
A: 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
$$\begin{align}
{1\over n^2}(1+\sqrt2+\cdots+\sqrt{n^2})
&\gt{1\over n^2}(1+1+1+2+2+2+2+2+3+\cdots+(n-1)+n)\\
&={1\over n^2}(3\cdot1+5\cdot2+7\cdot3+\cdots+(2n-1)(n-1)+n)\\
&\gt{2\over n^2}(1^2+2^2+3^2+\cdots+(n-1)^2)\\
&={2\over n^2}\cdot{(n-1)n(2n-1)\over6}\\
&={(n-1)(2n-1)\over3n}\\
&\to\infty
\end{align}$$
A: 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
$$\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) \ge \sqrt[2n]{n!}$$
A: Just another way assuming that you already heard about generalized hormonic numbers.
$$\sum_{k=1}^n \sqrt k=H_n^{\left(-\frac{1}{2}\right)}$$ Now, the asymptotics
$$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n^{3/2}}{3}+\frac{n^{1/2}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24n^{1/2}}+O\left(\frac{1}{n^{5/2}}
   \right)$$ So, for large values of $n$,
$$\frac{\sum_{k=1}^n \sqrt k } n=\frac{2 n^{1/2}}{3}+\frac{1}{2n^{1/2}}+O\left(\frac{1}{n}
   \right)$$ which answers the question but also gives an approximation.
Fo raxample, using $n=100$, the exact calculation would give $\approx 6.71463$ while the above formula would give $\frac{403}{60} \approx 6.71667$ (relative error $=0.03$%).
