TITLE: How to prove the inequality $2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$?
QUESTION [5 upvotes]: Prove that for any positive integer $n$,
$$2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$$
Progress
I think Riemann sum should be used for the middle term. I got the limit as $n\to \infty$ of the function inside using Riemann sum but what I cannot get are the lower and upper bound.

REPLY [2 votes]: The Riemann sum is not only useful in its limit when it approaches the integral.
If $f$ is an nonincreasing function, the lower and upper Riemann sum corresponding to a partitioning of $[1,n]$ at integer points are given by the left and right ends, i.e. we have
$$ \sum_{k=1}^{n-1} f(k)\le\int_1^{n}f(x)\,\mathrm dx\le \sum_{k=2}^{n} f(k)$$
Conversely, we conclude e.g.
$$ \int_1^{n}f(x)\,\mathrm dx+f(1)\le \sum_{k=1}^{n} f(k)\le\int_1^{n}f(x)\,\mathrm dx+f(n)$$
With $f(x)=\frac1{\sqrt x}$, the definite integral is $2\sqrt x\bigr|_1^n=2\sqrt n-2 $, so we get
$$ 2\sqrt n-1\le 1+\frac1{\sqrt 2}+\ldots+\frac1{\sqrt n}\le 2\sqrt n-2+\frac1{\sqrt n},$$
whichis slightly stronger than the problem statement asks for.