A sufficient condition for a recurrent sequence to have a limit Let $a_{n+3}=\left\{ a_{n}\right\} +\left\{ a_{n+1}\right\} +\left\{
a_{n+2}\right\} ,$ $\forall n\geq 1.$ How to find $a_{1},$ $a_{2},$ $%
a_{3}\in \mathbb{Q}\cap \left( 0,1\right) $, such that the sequence $\left(
a_{n}\right) _{n\geq 1}$ has a limit? ($\{x\}$ denotes the fractional part of $x$).
 A: Not a solution... just basic thoughts
Suppose that $\{a_n\}$ has a limit $l$. As the sequence takes values in $[0,3)$, $l \in [0,3]$.
As $x \mapsto \{x\}$ is continuous on $\mathbb R \setminus \mathbb Z$ if $l \notin \mathbb Z$ we must have $l=3\{l\}$. The only non integer solution is $3/2$.
If $l \in \mathbb Z$ then $l \in \{0,1,2,3\}$. Hence, $l$ has to belong to $\{0, 3/2, 1, 2, 3\}$.
And I have made no use of the hypothesis
$$H \equiv a_1, a_2, a_3 \in \mathbb Q \cap (0,1).$$
A: The only possible values of $\ a_1,a_2,a_3\ $ for which the sequence will converge are $\ a_1=a_2=a_3=\frac{1}{2}\ $.

*

*As Arthur observes in a comment on mathcounterexamples.net's observations the sequence can only converge if it's ultimately constant, and those observations show that the ultimate constant can only be $0,1,\frac{3}{2},$ or $2$.

*If the ultimate constant is an integer then since $\ a_{n+3}=\left\{a_n\right\}+\left\{a_{n+1}\right\}+\left\{a_{n+2}\right\}\ $, that integer can only be $0$, as Yves Daoust observes in a comment on mathcounterexamples.net's observations.

*If the limit were $0$, there would have to be a last integer $\ n\ $ for which $\ a_n\ne0\ $. Since $\ a_{n+3}=$$a_{n+2}=$$a_{n+1}=0\ $, $\ a_n\ $ must be an integer, as must $\ a_{n-1}\ $ and $\ a_{n-2}\ $. But then if $\ a_{n-3}\ $ were not an integer, then $\ a_n=\left\{a_{n-3}\right\}\ $ wouldn't be an integer either, and if $\ a_{n-3}\ $ were an integer, then $\ a_n\ $ would be $0$, which is a contradiction.  So the limit cannot be zero.

*Thus, the only possible limit is $\ \frac{3}{2}\ $. Again there must be a largest $\ n\ $ for which $\ a_n\ne\frac{3}{2}\ $.  The only possible values for $\ a_n\ $ are then $\ \frac{1}{2}\ $ or $\ \frac{5}{2}\ $ and the only possible values for $\ a_{n-1}\ $ and $\ a_{n-2}\ $ are $\ \frac{1}{2},\frac{3}{2}\ $ or $\ \frac{5}{2}\ $. But then $\ \left\{a_{n-3}\right\}+\left\{a_{n-2}\right\}+\left\{a_{n-1}\right\}=\left\{a_{n-3}\right\}+1<2\ $, so $\ a_n\ $ cannot be $\ \frac{5}{2}\ $, so must be $\ \frac{1}{2}\ $. But if $\ n>3\ $ then $ a_n=\left\{a_{n-3}\right\}+\left\{a_{n-2}\right\}+\left\{a_{n-1}\right\}=\left\{a_{n-3}\right\}+1>1\ $ so we must have $\ n\le3\ $.  Since $\ a_{n+1}=\frac{3}{2}>1\ $, however, we must have $\ n=3\ $. And, since $\ a_{n-1}=a_2<1\ $ and $\ a_{n-2}=a_1<1\ $ we must have $\ a_1=a_2=a_3=\frac{1}{2}\ $.

