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