Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$ 
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.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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
\\[5mm] & =
\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
\\[3mm] & = \bbx{\ds{\large\color{#f00}{60}}}
\end{align}

A: 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
$$\left\lfloor{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\right\rfloor=60$$
A: 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
$$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$
A: $$\sum_{k=2}^{1000}\frac1{\sqrt k}=\sum_{a=2}^{10}\frac1{\sqrt a}+\sum_{b=11}^{1000}\frac1{\sqrt b}$$
$$\sum_{a=2}^{10}\frac1{\sqrt a}\approx4.02100$$
Now notice that
$$\sqrt k-\sqrt{k-1}<\frac1{\sqrt k}<\sqrt{k+1}-\sqrt k$$
Which gives a telescoping series and...
$$56.64392\approx2(\sqrt{1001}-\sqrt{11})<\sum_{b=11}^{1000}\frac1{\sqrt b}<2(\sqrt{1000}-\sqrt{10})\approx56.92100$$
Thus,
$$60.66492<\sum_{k=2}^{1000}\frac1{\sqrt k}<60.94200$$
And so we see the solution is $60$.
