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 $$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With the identity $\ds{\sum_{k = 1}^{N}{1 \over n^{s}} = {N^{1 - s} \over 1 - s} + \zeta\pars{s} + s\int_{N}^{\infty}{\braces{x} \over x^{s + 1}}\,\dd x}$:
\begin{align} \left\lfloor\sum_{n = 2}^{1000}{1 \over \root{n}}\right\rfloor & = \left\lfloor-1 + \sum_{n = 1}^{1000}{1 \over \root{n}}\right\rfloor = \left\lfloor-1 + \pars{2\root{1000} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{1000}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x}\right\rfloor \\[5mm] & = \left\lfloor\underbrace{20\root{10} + \zeta\pars{1 \over 2} - 1} _{\ds{\approx\ \color{#f00}{60.7852}}}\ +\ \underbrace{{1 \over 2}\int_{1000}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x} _{\ds{\left\vert\begin{array}{l}\ds{> 0} \\ \mbox{and}\ <\ {\root{10} \over 100}\ \approx\ \color{#f00}{0.0316} \end{array}\right.}}\right\rfloor = \bbx{\ds{\large\color{#f00}{60}}} \end{align}