Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$ Prove that for any positive integer n,$$ 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots  + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 $$
Could anyone give me a hint on this question? Does it have something to do with Riemann Sums?

Hint. The double inequality
$$ 
2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ..  + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 
$$
can be proved inductively. For $n=1$ is clear. Assume that it holds for $n=k$. Then
$$ 
2\sqrt{k+1} - 2 +\frac{1}{\sqrt{k+1}}\le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ..  + \frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \le 2\sqrt{k} - 1 +\frac{1}{\sqrt{k+1}}.
$$
It remains to show the following two inequalities:
$$
2\sqrt{k+2} - 2\le 2\sqrt{k+1} - 2 +\frac{1}{\sqrt{k+1}},
$$
and
$$
2\sqrt{k} - 1 +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}-1.
$$
