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~

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.
