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

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

Thanks in advance!

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  • $\begingroup$ What do you mean by "with whole numbers?" The answer is not a whole number. Oh, just realized that the brackets may mean floor. $\endgroup$ – MCT Oct 6 '13 at 17:52
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    $\begingroup$ Does $[\cdot ]$ denote the greatest-integer / floor function (aka Gauß bracket)? $\endgroup$ – Daniel Fischer Oct 6 '13 at 17:52
  • $\begingroup$ Wolfram says the sum is $\approx 18.59$; do with the square brackets what you will. $\endgroup$ – vadim123 Oct 6 '13 at 17:57
  • $\begingroup$ @DanielFischer Yes the [] structure denotes the floor function. Say x is 3.14 than [x] is 3. $\endgroup$ – Dora Benzo Oct 6 '13 at 18:05
  • $\begingroup$ @vadim123 Thanks for your response but I also need a mathematical proof. :) $\endgroup$ – Dora Benzo Oct 6 '13 at 18:07
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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 $$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}} &< 1 + (2 \sqrt{2} - 2 \sqrt{1}) + \ldots + (2 \sqrt{100} - 2 \sqrt{99})\cr & = 1 + 20 - 2 = 19\cr}$$

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    $\begingroup$ It seem that your solution solves my problem. However I was wondering what was your process of thinking which led to that 1/sqrt(t) + 2*sqrt(t) assumption. I would have never guessed it. Thanks again! $\endgroup$ – Dora Benzo Oct 6 '13 at 19:06
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    $\begingroup$ The idea came from the integral... $\endgroup$ – Robert Israel Oct 6 '13 at 19:13
  • $\begingroup$ ok...I'm thinking at my theacher..he may asks me why I solved in this way the problem. $\endgroup$ – Dora Benzo Oct 6 '13 at 19:28
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    $\begingroup$ Why not be honest and tell him that you got help here? $\endgroup$ – Robert Israel Oct 6 '13 at 21:26
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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.

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  • $\begingroup$ Thanks again for your involvement. The trickier part is that the exercise should be resolved using 9th grade math. This means that no mathematical analysis concept should be used. $\endgroup$ – Dora Benzo Oct 6 '13 at 18:15
  • $\begingroup$ Very Nice Solution vadim123, Using Integration. $\endgroup$ – juantheron Apr 14 '15 at 17:02

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