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!
 A: 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.
A: 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
$$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}}
&< 1 + (2 \sqrt{2} - 2 \sqrt{1}) + \ldots + (2 \sqrt{100} - 2 \sqrt{99})\cr
& = 1 + 20 - 2 = 19\cr}$$
