Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$
This is all about providing an accurate approximation for the involved generalized harmonic sum. I will use a technique (creative telescoping) clearly outlined in the first chapter of these course notes.
We may notice that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n-\frac{1}{2}}} $$
is a telescopic term and it is, additionally, pretty close to $\frac{1}{2\sqrt{n}}$. In particular
$$ \frac{1}{\sqrt{n}}-2\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right)= d_n\\=-\frac{1}{2\sqrt{n}\left(\sqrt{n+1/2}+\sqrt{n-1/2}\right)^2\left(\sqrt{n+1/2}+\sqrt{n}\right)\left(\sqrt{n-1/2}+\sqrt{n}\right)}$$
is a negative term that behaves like $-\frac{1}{32 n^{5/2}}$ for large $n$s. It follows that
$$ \sum_{k=2}^{1000}\frac{1}{\sqrt{k}} = 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)+\sum_{k=2}^{1000}d_k $$
has a distance from
$$ 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right) = 2\left(\sqrt{1000+\frac{1}{2}}-\sqrt{2-\frac{1}{2}}\right) $$
that is less$^{(*)}$ than $\frac{8}{1000}$. By computing the last quantity it follows that the answer is $\color{red}{60}$.
$(*)$ Proof: we have $$ |d_k|\leq \frac{1}{48 k^{3/2}}-\frac{1}{48(k+1)^{3/2}} $$ hence $$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$
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
$$\left\lfloor{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\right\rfloor=60$$