Find sum of the series

$\begin{aligned}S&=\sum_{x=1}^{100} \dfrac{1}{\sqrt{x}}\\ \text{We have, } \\ \dfrac{2}{\sqrt{x}+\sqrt{x}}&<\dfrac{2}{\sqrt{x}+\sqrt {x-1}} \hspace{4mm}\color{#3D99F6} x\geq1\\ \implies \dfrac{1}{\sqrt{x}}&<2\times(\sqrt{x}-\sqrt{x-1})\\ \implies 1 +\sum_{x=2}^{100} \dfrac{1}{\sqrt{x}}&<1+\sum_{x=2}^{100} 2\times(\sqrt{x}-\sqrt{x-1})\\ \implies \sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}&<1+2(\sqrt{100}-\sqrt{1})\\ \implies \sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}&<19 \hspace{4mm}\color{#3D99F6} (1)\\\\ \text{Similarly,}\\ \dfrac{2}{\sqrt{x}+\sqrt{x}}&>\dfrac{2}{\sqrt{x}+\sqrt {x+1}}\\ \implies \dfrac{1}{\sqrt{x}}&>2\times(\sqrt{x+1}-\sqrt{x})\\ \implies \sum_{x=1}^{100} \dfrac{1}{\sqrt{x}}&>\sum_{x=1}^{100} 2\times(\sqrt{x+1}-\sqrt{x})\\ \implies \sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}&>2(\sqrt{101}-\sqrt{1})>2(\sqrt{100}-\sqrt{1})\\ \implies \sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}&>18 \hspace{4mm}\color{#3D99F6} (2)\\\\ \text{From } \color{#3D99F6}(1) \color{#333333}\text{ and } \color{#3D99F6}(2),\\ 18&<\sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}<19 \\ \implies \Bigl \lfloor{\sum_{x=1}^{100}\dfrac{1}{\sqrt{x}}}\Bigr \rfloor&=18\end{aligned}$

We know the exact values of $1/\sqrt{4}$ , $1/\sqrt{9}$ , etc. The neighboring terms have similar values. Divide them into "chunks": $\frac1{\sqrt3} + \frac1{\sqrt4} + \frac1{\sqrt5} + \frac1{\sqrt6} \approx \frac 4 2 = 2; \\ \frac 1{\sqrt7} + \frac 1{\sqrt8} + \frac 1{\sqrt9} + \frac 1{\sqrt{10}} + \frac 1{\sqrt{11}} + \frac 1{\sqrt{12}} \approx \frac 6 3 = 2 \\ \vdots \\ \frac 1{\sqrt{73}} + \cdots + \frac 1{\sqrt{90}} \approx \frac{18}{9} = 2$ $\frac 1{\sqrt{91}} + \cdots + \frac 1{\sqrt{100}} \approx \frac{10}{10} = 1$ The sums of each of these 9 terms is approximately equal to 17. Add the first two terms, $1 + 1/\sqrt{2} \approx 1 + \tfrac32$ to conclude that the sum is close to $18\tfrac12$ .

We begin by knowing the fact that $\displaystyle f(x)=\dfrac{1}{\sqrt{x}}$ is decreasing and also by drawing some rectangles to estimate the integral we have,

$\displaystyle s-1\leq \int_{1}^{100} {\dfrac{1}{\sqrt{x}}dx} \leq s$

So,

$\displaystyle 18 \leq s \leq 18.17$ and therefore,

$\displaystyle \left\lfloor s \right\rfloor =\boxed{18}$