How to find $\lfloor 1/\sqrt{1}+1/\sqrt{2}+\dots+1/\sqrt{100}\rfloor $ without a calculator?

$$ \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? $$

I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.

Thanks in advance!

Doing it in 9th grade math is quite a challenge. But perhaps this would come close.

For any positive number $t$, we have $$ \dfrac{1}{\sqrt{t}} > 2 \sqrt{t+1} - 2 \sqrt{t} > \dfrac{1}{\sqrt{t+1}}$$

To see the first inequality, note that

$$ \left(\dfrac{1}{\sqrt{t}} + 2 \sqrt{t}\right)^2 = \dfrac{1}{t} + 4 + 4 t > 4 + 4 t = (2 \sqrt{t+1})^2$$ Similarly for the second, by looking at $\left(2 \sqrt{t+1} - \dfrac{1}{\sqrt{t+1}}\right)^2$.

So $$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}} &> (2 \sqrt{2} - 2 \sqrt{1}) + (2 \sqrt{3} - 2 \sqrt{2}) + \ldots + (2 \sqrt{101} - 2 \sqrt{100})\cr &= 2 \sqrt{101} - 2 > 2 \sqrt{100} - 2 = 18\cr}$$ while $$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}} &< 1 + (2 \sqrt{2} - 2 \sqrt{1}) + \ldots + (2 \sqrt{100} - 2 \sqrt{99})\cr & = 1 + 20 - 2 = 19\cr}$$

You can get a lower bound by $$\sum_{i=1}^{100}\frac{1}{\sqrt{i}}> \int_1^{101}\frac{dx}{\sqrt{x}}=2\sqrt{101}-2\approx 18.10$$

You can get an upper bound by $$\sum_{i=1}^{100}\frac{1}{\sqrt{i}}< \int_0^{100}\frac{dx}{\sqrt{x}}=20$$

We can refine that upper bound by replacing $\int_0^1\frac{dx}{\sqrt{x}}$ with $1$, which replaces 20 with 19. Hence the desired sum lies between 18 and 19, with floor 18.