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$$A = \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10000}}$$

Find $\lfloor A\rfloor$ where $\lfloor x\rfloor$ is the greatest integer less than, or equal to $x$

I got stuck on this, so when I finally did it, I decided to post it here. And of course, I am always looking for alternatives, so keep answering.

Guy
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  • Related post : http://math.stackexchange.com/questions/465857/sum-of-the-series-sum-i-i-alpha/465868#465868 – user88595 Mar 16 '14 at 19:30
  • @user88595 it is almost intractable for unknown alpha though. but thanks for linking. – Guy Mar 16 '14 at 19:32
  • Where did you get this problem? I remember seeing something similar (or identical) on brilliant.org – MT_ Mar 16 '14 at 20:00
  • @user92774 friend gave it to me. said it was on in an olympiad or something. – Guy Mar 17 '14 at 04:21
  • @Sabyasachi Cool deal. Makes sense for it to be on an olympiad since it is very easy to guess on using $\int_{2}^{10000} \frac{1}{x} dx$, but hard to show that that estimation is accurate. – MT_ Mar 17 '14 at 04:27
  • @user92774 true. although you meant $\int_{2}^{10000}\frac{1}{\sqrt{x}},dx$ I believe? I tried that, wasn't tight. accurate estimation by that gives $196\lt A\lt199$. Right, but not helpful. – Guy Mar 17 '14 at 04:29
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    You could obtain a much tighter overestimate by considering the given sum as a midpoint-rule Riemann sum for $\int_{2-1/2}^{10000+1/2} \frac{dx}{\sqrt{x}} \approx 197.555510.$ Of course obtaining the decimal representation isn't as easy here :-) – Antonio Vargas Mar 17 '14 at 12:30
  • @AntonioVargas and the actual sum is $197.544$. nice. – Guy Mar 17 '14 at 12:32

2 Answers2

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We start by noting that, $$\frac{1}{\sqrt{k}}=\frac2{\sqrt{k}+\sqrt{k}}\lt\frac2{\sqrt{k}+\sqrt{k-1}}$$

So we have, $$\frac{1}{\sqrt{k}}\lt \frac{2}{\sqrt{k}+\sqrt{k-1}} = 2(\sqrt{k} - \sqrt{k-1})$$

Thus we have $$S=\sum_{i=2}^{10000}\frac{1}{\sqrt{k}} \lt \sum_{i=2}^{10000}2(\sqrt{k} - \sqrt{k-1}) = 2(\sqrt{10000}-\sqrt1) = 198$$

$$\color{red}{A\lt198}\tag{1}$$

Also,

$$\frac{1}{\sqrt{k}}=\frac2{\sqrt{k}+\sqrt{k}}\gt\frac2{\sqrt{k}+\sqrt{k+1}}$$

And so,

$$\frac{1}{\sqrt{k}}\gt \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1} - \sqrt{k})$$

And therefore,

$$S=\sum_{i=2}^{10000}\frac{1}{\sqrt{k}} \gt \sum_{i=2}^{10000}2(\sqrt{k+1} - \sqrt{k}) = 2(\sqrt{10001}-\sqrt2) \gt 197$$

$$\color{red}{A\gt197}\tag{2}$$

Combining $(1)$ and $(2)$

$$197\lt A \lt 198$$

$$\implies \lfloor A\rfloor = 197$$

Guy
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    It's amusing that $\displaystyle{\large\int_{2}^{10000}{\dd x \over ,\sqrt{,x,},} = 200 - 2,\sqrt{,2,}, = 197.17\ldots}$. – Felix Marin Mar 16 '14 at 19:44
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$$ \begin{align} 197 &< 2\left(\sqrt{10001}-\sqrt{2}\right) \\ &= \int_2^{10001} \frac{dx}{\sqrt{x}} \\ &< \color{blue}{\sum_{k=2}^{10000} \frac{1}{\sqrt{k}}} \\ &< \int_1^{10000} \frac{dx}{\sqrt{x}} \\ &= 198. \end{align} $$

Antonio Vargas
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