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~

For the right inequality:
$$\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) = \frac{1}{n}\left(\sqrt{n}+\sqrt{\frac{n}{2}}+\cdots+\sqrt{\frac{n-1}{n}} + \sqrt{\frac{n}{n}}\right)$$
This is a Riemann sum of a function $f(x) = \frac{1}{\sqrt{x}}$ on the segment $[0,1]$ with the equidistant subdivision with endpoints $\left\{0, \frac{1}{n}, \frac{2}{n}, \ldots, 1\right\}$.
So:
$$\lim_{n\to\infty} \frac{1}{n}\left(\sqrt{n}+\sqrt{\frac{n}{2}}+\cdots+ \sqrt{\frac{n}{n}}\right) = \int_{0}^1 \frac{dx}{\sqrt{x}} = 2$$
Since the Riemann sums increase towards the integral, by multiplying with $\sqrt{n}$ we obtain the desired inequality.
