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Given a sequence, $a_n=|a_{n-1}-a_{n-2}|$.

for any squence $\{a_n\}$ like this, how can I prove the existence of $k \in \mathbb{N}$ such that $0\le a_k<1$?

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    $\begingroup$ It is not clear what you are asking $\endgroup$ – Yiorgos S. Smyrlis May 11 '17 at 7:20
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    $\begingroup$ What is $N$ supposed to be? $\endgroup$ – infinitylord May 11 '17 at 7:33
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    $\begingroup$ Isn't $0,4,4,0,4,4,0,4,4,0,\cdots$ and example of such a sequence? Then you won't have $a_n<1$ eventually (if that's what you are aiming at). $\endgroup$ – drhab May 11 '17 at 7:42
  • $\begingroup$ not for any n,just one $a_n$ is enough. $\endgroup$ – Charles Bao May 11 '17 at 7:46
  • $\begingroup$ @infinitylord natrual number set $\endgroup$ – Charles Bao Oct 10 '17 at 1:35
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We may suppose that all $a_n$ are $\geq 0$. Put $u_n=\max\{a_n,a_{n+1}\}$. We have $a_{n+1}\leq u_n$ and $a_{n+2}=|a_{n+1}-a_n|\leq u_n$, hence $u_{n+1}\leq u_n$. Now $2u_n=a_n+a_{n+1}+|a_{n+1}-a_n|=a_n+a_{n+1}+a_{n+2}$; we get that $a_{n+3}\leq a_n$. This imply that the sequences $a_{3k+r}$, $r=0,1,2$ are decreasing and hence have a limit $L_r\geq 0$. Now using the recurrence relation we get $L_0=|L_1-L_2|$, $L_1=|L_0-L_2|$, $L_2=|L_1-L_0|$. If we have $u,v,w$ such that $u=|v-w|$, $v=|u-w|$, $w=|v-u|$, we may suppose that $u\geq v\geq w$, this gives that $u=v-w$ , $v=u-w$ and $w=v-u$, and this imply that $w=0$ and $u=v$. Hence there exist $r_0$ such that $L_{r_0}=0$, and for $k$ large, we have $a_{3k+r_0}<1$.

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  • $\begingroup$ how to prove $a_{n+3}\leq a_n$? $\endgroup$ – Charles Bao May 11 '17 at 8:19
  • $\begingroup$ @Charles Bao Write $u_{n+1}\leq u_n$ and simplify (using $2u_n=a_n+a_{n+1}+a_{n+2}$) $\endgroup$ – Kelenner May 11 '17 at 8:20

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