Prove that for any positive integer $n$, $$2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$$
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.
On
There's a tricky solution using integral estimates: $$2\sqrt{n+1}-2=\int_1^{n+1}\frac1{\sqrt n}\le\sum_{k=1}^n\frac1{\sqrt n}=1+\sum_{k=2}^n\frac1{\sqrt n}\le 1+\int_1^n\frac1{\sqrt {n}}=1+2\sqrt n-2=2\sqrt n-1$$
But you can easily proceed by induction, the base case is trivial and $$(2\sqrt{n+1}-1)-(2\sqrt n-1)=2(\sqrt{n+1}-\sqrt n)=2\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n}=\frac2{\sqrt{n+1}+\sqrt n}\ge\frac1{\sqrt{n+1}}$$ So the difference of successive right hand sides is bigger than what you add to the middle.
And similarly the other inequality: $$(2\sqrt{n+2}-2)-(2\sqrt{n+1}-2)=2(\sqrt{n+2}-\sqrt{n+1})=\frac2{\sqrt{n+2}+\sqrt{n+1}}\le\frac1{\sqrt{n+1}}$$
On
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.
Hint: Notice $\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}<\dfrac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1}),\quad n>1$
Similarly, $\dfrac{1}{\sqrt{n}}>2(\sqrt{n+1}-\sqrt{n}),\quad n>1$