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This question already has an answer here:

Let $$ P = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{4}} ... +\frac{1}{\sqrt{10000}}$$ what is the value of the floor function of P?

My try:

I tried assuming these 2 bounds

$$ P_x = 1 + 1 + 1 + \frac{1}{2}+...\frac{1}{99 }$$ where it is repeated until the next square number (eg. there are 3 1's at the beginning of the sequence corresponding to the $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{4}}$ where $\frac{1}{\sqrt{4}}$ is the next square number

and

$$ P_y = \frac{1}{2} + \frac{1}{2} +\frac{1}{2} + \frac{1}{3} + \frac{1}{3} ... \frac{1}{100}$$ withe the same counting process as $P_x$

then we know that

$$P_x>P>P_y$$

$$99*2 + (\frac{1}{1} + \frac{1}{3} + \frac{1}{4} ... \frac{1}{99}) > P >99*2 -(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} ... \frac{1}{100}) $$

but as you can see, the floor function of P can be either 197 or 198, how would I answer this?

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marked as duplicate by Martin R, Paul Frost, Sil, rtybase, Robert Z sequences-and-series Mar 24 at 19:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You can use the following inequality. $$\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-\sqrt{k}).$$ We obtain: $$\sum_{k=2}^{10000}\frac{1}{\sqrt{k}}>2\sum_{k=2}^{10000}(\sqrt{k+1}-\sqrt{k})=2(\sqrt{10001}-\sqrt2)>197.$$ Also, we have $$\frac{1}{\sqrt{k}}<2(\sqrt{k}-\sqrt{k-1}).$$

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  • $\begingroup$ what happened to the 2nd expression in the string ? if possible though, a solution without calculus would be nice $\endgroup$ – SuperMage1 Mar 24 at 7:13
  • $\begingroup$ @SuperMage1 It was typo. I fixed. $\endgroup$ – Michael Rozenberg Mar 24 at 7:15
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From the definition of the Riemann integral we can say:

$\sum_\limits{n=2}^{10000} \frac {1}{\sqrt n} \le \int_1^{10000}\frac {1}{x^\frac 12}\ dx \le\sum_\limits{n=1}^{9999} \frac {1}{\sqrt n}$

or

$\int_2^{10001}\frac {1}{x^\frac 12}\ dx \le \sum_\limits{n=2}^{10000} \frac {1}{\sqrt n} \le \int_1^{10000}\frac {1}{x^\frac 12}\ dx $

$2 (\sqrt {10001} - \sqrt 2)\le\sum_\limits{n=2}^{1000} \frac {1}{\sqrt n} \le 2 (\sqrt {10000} - 1)$

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  • $\begingroup$ It is Georg Friedrich Bernhard Riemann, not Reimann :) $\endgroup$ – Martin R Mar 24 at 7:42
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Use $$\int_2^{10001}\frac{\mathrm{d}x}{\sqrt{x}}<P<\int_1^{10000}\frac{\mathrm{d}x}{\sqrt{x}}.$$

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