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Recurrent Sequence implies Constant


Prove that the sequence which satisfies the relation a_{n+1}=\sqrt{6-2a_n^2},\ \forall n\geq 1 is constant.
Laurentiu Panaitopol, Romania
Solution: We can translate the recurrence relation in the following way:
a_{n+1}^2+2a_n^2=6,\ (0). Therefore a_{n+2}^2+2a_{n+1}^2=6=a_{n+1}^2+2a_n^2, which implies a_{n+2}^2-a_{n+1}^2=(-2)(a_{n+1}^2-a_n^2),\ (1).
From (0) it follows that (a_n) is bounded, and from (1) it follows that y_n=a_{n+1}^2-a_n^2 is unbounded if it is not zero. This would contradict the fact that (a_n) is bounded. Therefore (y_n) is bounded and (a_n^2) is constant. Because a_n \geq 0,\ \forall n \geq 2 we get that (a_n) is constant.

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