Prove identity holds Just wondering if there is any other way to show that for each positive integer $n$ holds
$$2\left(\sqrt{n} - 1\right) < 1 + \frac{1}{\sqrt{2}} +\cdots+\frac{1}{\sqrt{n}} < 2\sqrt{n}$$
other than by mathematical induction~

Note that
$$\frac{1}{k+1}<\int_k^{k+1}\frac{dx}{\sqrt{x}}<\frac{1}{k}$$
for every $k\geq0$. Adding this relations we get
$$\int_1^{n+1}\frac{dx}{\sqrt{x}}<1 + \frac{1}{\sqrt{2}} +\cdots+\frac{1}{\sqrt{n}}<\int_0^n\frac{dx}{\sqrt{x}}$$
Solving the integrals we get what we wanted.
