Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$

Question

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?

Answer 1

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 $$ 2\sqrt{k} - 1 +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}-1. $$

Answer 2

Hints:

$$2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k}+\sqrt{k+1}} < \frac{1}{\sqrt{k}}< \frac{2}{\sqrt{k-1}+\sqrt{k}}=2(\sqrt{k}-\sqrt{k-1})$$