$2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$ If $n\in N$,then prove that $2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$
How should i prove this inequality,neither AM-HM nor any other inequality theorem is working here.
 A: Denote the sum by $S_n$. Looking at the graph of $f(x)=\frac{1}{\sqrt{x}}$ it is clear that you have $$S_n <1+\int_1^{n} \frac{1}{\sqrt{x}} dx=1+[2\sqrt{x}]_1^n=2\sqrt{n}-1$$
Can you also derive the lower bound by a slight variation of this method?
A: This inequality comes in hand: $$ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} < \sqrt{n} - \sqrt{n-1} $$
Part 1: $ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} $
$$ \sqrt{n+1} - \sqrt{n} = (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$
Since $\sqrt{n+1} > \sqrt{n} $, then $\, \sqrt{n+1} + \sqrt{n} > 2 \sqrt{n} \implies \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}}$
So: $$ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$$
Part 2: $ \sqrt{n} - \sqrt{n-1} > \frac{1}{2\sqrt{n}}$
$$ \sqrt{n} - \sqrt{n-1} = (\sqrt{n} - \sqrt{n-1}) \times \frac{\sqrt{n} + \sqrt{n-1}}{\sqrt{n} + \sqrt{n-1}} = \frac{1}{\sqrt{n} + \sqrt{n-1}}$$
Since $\sqrt{n-1} < \sqrt{n} \implies \sqrt{n-1} + \sqrt{n} < 2 \sqrt{n} \implies \frac{1}{\sqrt{n} + \sqrt{n-1}} > \frac{1}{2\sqrt{n}} $
So: $$\sqrt{n} - \sqrt{n-1} > \frac{1}{2\sqrt{n}} $$
Now you have the following inequalities:
$$\sqrt{2} - \sqrt{1} < \frac{1}{2\sqrt{1}} < \sqrt{1} - \sqrt{0} $$
$$\sqrt{3} - \sqrt{2} < \frac{1}{2\sqrt{2}} < \sqrt{2} - \sqrt{1} $$
$$\sqrt{4} - \sqrt{3} < \frac{1}{2\sqrt{3}} < \sqrt{3} - \sqrt{2} $$
$$ \ldots $$
