For natural number $n$ we define \[ a_n = \{ \sqrt{n} \} - \{ \sqrt{n + 1} \} + \{ \sqrt{n + 2} \} - \{ \sqrt{n + 3} \}. \] a) Show that $a_1 > 0,2$. b) Show that $a_n < 0$ for infinity many values of $n$ and $a_n > 0$ for infinity values of natural numbers of $n$ as well. ( We denote by $\{ x \} $ the fractional part of $x.$)
Problem
Source: Romania National Olympiad 2023
Tags: fractional part, algebra, Inequality
zjugh
23.06.2023 06:32
(1)$a_1=\{\sqrt 3\}-\{\sqrt 2\}=\sqrt 3-\sqrt 2$,notice that $\frac 1{a_1}=\sqrt 2+\sqrt 3<5$,$\therefore a_1>0.2$. (2)if $n=k^2-3(k\in \mathbb{N},k\ge 3)$,then $\{\sqrt {n+3}\}=0$,thus $a_n=\{\sqrt n\}+\sqrt {n+2}-\sqrt {n+1}>0$,s so $a_n>0$ for infinity many values of $n$; if $n=k^2(k\in \mathbb{N},k\ge 3)$,then $\{\sqrt n\}=0$,so $a_n=-\{\sqrt {n+1}\}+\{\sqrt {n+2}\}-\{\sqrt {n+3}\}<0$. so $a_n<0$ for infinity many values of $n$.
David_Kim_0202
23.06.2023 07:48
a) is trivial if you put $1$ into $n$ b) For every square number $p$ $$a_p=-\{\sqrt{p+1}\}+\{\sqrt{p+2}\}-\{\sqrt{p+3}\}$$Because, $\{\sqrt{p+2}\}-\{\sqrt{p+3}\}$ is a negative number, for all square number $p$ $a_n$ is negative. Also, for all $a, b (a>b)$, $$\sqrt{a}-\sqrt{a-1}>\sqrt{b}-\sqrt{b-1}$$so if $n=x^2-3$ this works.
lpieleanu
01.07.2023 20:32
We have $$a_1 = \{ \sqrt{1} \} - \{ \sqrt{2} \} + \{ \sqrt{3} \} - \{ \sqrt{4} \} = (1-\sqrt{2}) + (\sqrt{3}-1) = \sqrt{3} - \sqrt{2}.$$Therefore, it suffices to prove that $$\sqrt{3} - \sqrt{2} > 0.2.$$Note that $(2.48)^2 = 6.1504 > 6,$ so $2.48 > \sqrt{6}.$ Therefore, it follows that $$5 - 2\sqrt{6} > 0.04.$$Taking the square root of both sides yields $$\sqrt{3} - \sqrt{2} > 0.2,$$so we are done. $\square$
Proof for >0: Obviously, there are an infinite number of natural numbers $n$ such that $$\lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+1} \rfloor = \lfloor \sqrt{n+2} \rfloor = \lfloor \sqrt{n+3}\rfloor.$$In these cases, we have $$a_n = \{ \sqrt{n} \} - \{ \sqrt{n + 1} \} + \{ \sqrt{n + 2} \} - \{ \sqrt{n + 3} \}= \sqrt{n} - \sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}.$$Note that $$n^2 + 3n < n^2 + 3n + 2,$$for all $n \in \mathbb{Z}_{\geq 1}.$ Manipulating this yields $$\sqrt{n^2 + 3n} < \sqrt{n^2+3n+2}$$$$-2\sqrt{n^2+3n} > -2\sqrt{n^2+3n+2}$$$$n+3 - 2\sqrt{n^2+3n} + n > n+2 - 2\sqrt{n^2+3n+2} + n+1$$$$(\sqrt{n+3} - \sqrt{n})^2 > (\sqrt{n+2} - \sqrt{n+1})^2$$$$(\sqrt{n} - \sqrt{n+3})^2 > (\sqrt{n+1} - \sqrt{n+2})^2$$$$\sqrt{n} - \sqrt{n+3} > \sqrt{n+1} - \sqrt{n+2}$$$$\sqrt{n} - \sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}>0,$$so we are done with this part.
Proof for <0: Let $n$ be a perfect square such that none of the rest of the numbers are perfect squares. (Obviously, there is an infinite number of such constructions.) Then, we have $$a_n = -\sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}.$$Clearly, $n+2 < n+3,$ so we have $$\sqrt{n+2} < \sqrt{n+3}$$which means that $a_n$ must be negative, as desired. $\square$