TITLE: Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$
QUESTION [11 upvotes]: Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.

I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me.

REPLY [2 votes]: The inequality
$${1\over\sqrt n}\gt\int_n^{n+1}{dx\over\sqrt x}=2(\sqrt{n+1}-\sqrt n)$$
is geometrically obvious, since $1\over\sqrt x$ is a decreasing function.  The inequality
$${1\over\sqrt n}\lt\int_{n-{1\over2}}^{n+{1\over2}}{dx\over\sqrt x}=2\left(\sqrt{n+{1\over2}}-\sqrt{n-{1\over2}}\right)$$
is not obvious, but straightforward to verify algebraically:
$$\begin{align}
{1\over\sqrt n}\lt2\left(\sqrt{n+{1\over2}}-\sqrt{n-{1\over2}}\right)
&\iff{1\over\sqrt{2n}}\lt\sqrt{2n+1}-\sqrt{2n-1}\\
&\iff{1\over2n}\lt(2n+1)-2\sqrt{4n^2-1}+(2n-1)\\
&\iff\sqrt{4n^2-1}\lt2n-{1\over4n}\\
&\iff4n^2-1\lt4n^2-1+{1\over16n^2}
\end{align}$$
We therefore get
$$\int_2^{1001}{dx\over\sqrt x}\lt{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\lt\int_{3/2}^{2001/2}{dx\over\sqrt x}$$
From $\int_2^{1001}{dx\over\sqrt x}=2(\sqrt{1001}-\sqrt2)\approx60.4487$ and $\int_{3/2}^{2001/2}{dx\over\sqrt x}=2(\sqrt{2001/2}-\sqrt{3/2})\approx60.81187$ we find
$$\left\lfloor{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\right\rfloor=60$$