$2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$

Question

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.

Answer 1

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?

Answer 2

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:

$$\sqrt{2} - \sqrt{1} < \frac{1}{2\sqrt{1}} < \sqrt{1} - \sqrt{0} $$ $$\sqrt{3} - \sqrt{2} < \frac{1}{2\sqrt{2}} < \sqrt{2} - \sqrt{1} $$ $$\sqrt{4} - \sqrt{3} < \frac{1}{2\sqrt{3}} < \sqrt{3} - \sqrt{2} $$ $$ \ldots $$