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~
 A: 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.
A: $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality
$$ \frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\leq\int_{1/2}^{n+1/2}\frac{dx}{\sqrt{x}}=\sqrt{4n+2}-\sqrt{2}\leq 2\sqrt{n}-1 $$
as well as
$$ \frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\geq \frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{n}}+\int_{1}^{n}\frac{dx}{\sqrt{x}} \geq 2\sqrt{n}-\tfrac{3}{2}. $$
A: 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.
A: Consider the graph of $f(x)=\frac{1}{\sqrt{x}}$.  Observe that this is a decreasing function.
For the upper bound: Divide up the interval $[0,n]$ into $n$ subintervals of size $1$.  Now, consider the rectangles of the form $[a,a+1]\times[0,\frac{1}{\sqrt{a+1}}]$.  The area of each rectangle is $\frac{1}{\sqrt{a+1}}$.  Therefore, the sum of the areas of these rectangles is exactly the sum in the middle.  On the other hand, the area of these rectangles is less than the area under the curve $f(x)$.  Since
$$
\int_0^n\frac{dx}{\sqrt{x}}=\left.2\sqrt{x}\right|_0^n=2\sqrt{n},
$$ 
the result follows.
For the lower bound: This is similar, but now we want to compare to $\int_1^n\frac{dx}{\sqrt{x}}$.  In this case, one would divide up the interval $[1,n+1]$ into $n$ subintervals of size $1$.  Now, consider the rectangles of the form $[a,a+1]\times[0,\frac{1}{\sqrt{a}}]$ the area of each rectangle is $\frac{1}{\sqrt{a}}$.  Therefore the sum of the areas of these rectangles is exactly the sum on the middle.  On the other hand, the area of these rectangles is greater than the area under the curve $f(x)$.  Since
$$
\int_1^{n+1}\frac{dx}{\sqrt{x}}=2\sqrt{x}|_1^{n+1}=2\sqrt{n+1}-2,
$$
and $\sqrt{n}\leq \sqrt{n+1}$, the result follows.
A: The outcome of the standard sum/integral comparison is
$$  2 \sqrt {n+1} - 2 \;  < \; 1  + \frac{1}{\sqrt 2 }  + \frac{1}{\sqrt 3 } + \cdots + \frac{1}{\sqrt n } \; < \; 2 \sqrt n - 1 $$

A: Upper Bound:
$$H_k=1+\sum_{k=2}^n\frac{1}{n}\leq 1+\sum_{k=2}^n\frac{1}{\sqrt{k(k-1)}}\leq^{CS} 1+\sqrt{(n-1)\sum_{k=2}^n\frac{1}{k(k-1)}}\leq 1+\sqrt{(n-1)  \left(\frac{1}{2}-\frac{1}{n}\right)}$$
Which is even better than yours (CS stands for Cauchy-Schwarz).
For the lower bound, work on:
$$\sqrt{k-1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\approx \frac{2}{\sqrt{k}}$$
