6
$\begingroup$

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

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.

$\endgroup$
8
  • $\begingroup$ Related post : math.stackexchange.com/questions/465857/… $\endgroup$
    – user88595
    Mar 16 '14 at 19:30
  • $\begingroup$ @user88595 it is almost intractable for unknown alpha though. but thanks for linking. $\endgroup$
    – Guy
    Mar 16 '14 at 19:32
  • $\begingroup$ Where did you get this problem? I remember seeing something similar (or identical) on brilliant.org $\endgroup$
    – MCT
    Mar 16 '14 at 20:00
  • $\begingroup$ @user92774 friend gave it to me. said it was on in an olympiad or something. $\endgroup$
    – Guy
    Mar 17 '14 at 4:21
  • 1
    $\begingroup$ 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 :-) $\endgroup$ Mar 17 '14 at 12:30
14
$\begingroup$

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 $(2)$

$$197\lt A \lt 198$$

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

$\endgroup$
1
  • 1
    $\begingroup$ It's amusing that $\displaystyle{\large\int_{2}^{10000}{\dd x \over \,\sqrt{\,x\,}\,} = 200 - 2\,\sqrt{\,2\,}\, = 197.17\ldots}$. $\endgroup$ Mar 16 '14 at 19:44
12
$\begingroup$

$$ \begin{align} 197 &< 2\left(\sqrt{10001}-\sqrt{2}\right) \\ &= \int_2^{10001} \frac{dx}{\sqrt{x}} \\ &< \color{blue}{\sum_{k=2}^{10000} \frac{1}{\sqrt{k}}} \\ &< \int_1^{10000} \frac{dx}{\sqrt{x}} \\ &= 198. \end{align} $$

$\endgroup$
1
  • $\begingroup$ nice. +1${}{}{}$ $\endgroup$
    – Guy
    Mar 17 '14 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.