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Romanian Mathematical Olympiad

Algebra Difficulty 5.8 AIME, harder Prove it Romania

Let α\alpha be an irrational number. For any n∈N∗n \in \mathbb{N}^* let an={nα}a_n = \{n\alpha\} and define the sequence (xn)n≥1(x_n)_{n \ge 1} by xn=(a2−a1)(a3−a2)⋯(an+1−an)x_n = (a_2 - a_1)(a_3 - a_2) \cdots (a_{n+1} - a_n). Show that the sequence is convergent and find its limit.

Solution

We claim that {x+y}−{y}\{x + y\} - \{y\} is equal to {x}\{x\} or {x}−1\{x\} - 1, for any real numbers xx and yy. Write x+y=[y]+[x]+{x}+{y}x + y = [y] + [x] + \{x\} + \{y\} to get {x+y}−{y}={{x}+{y}}−{y}\{x + y\} - \{y\} = \{\{x\} + \{y\}\} - \{y\}. Hence if {x}+{y}<1\{x\} + \{y\} < 1 then {x+y}−{y}={x}\{x + y\} - \{y\} = \{x\}, while if 2>{x}+{y}≥12 > \{x\} + \{y\} \ge 1 then {x+y}−{y}={x}−1\{x + y\} - \{y\} = \{x\} - 1, as claimed.

Apply the above result to infer that either ∣an+1−an∣={α}|a_{n+1} - a_n| = \{\alpha\} or ∣an+1−an∣=1−{α}|a_{n+1} - a_n| = 1 - \{\alpha\}. Set b=max⁡{{α},1−{α}}b = \max\{\{\alpha\}, 1 - \{\alpha\}\} and notice that 0<b<10 < b < 1 (for α\alpha is irrational) to derive that ∣xn∣≤bn|x_n| \le b^n, implying lim⁡n→∞∣xn∣=0\lim_{n \to \infty} |x_n| = 0 and furthermore lim⁡n→∞xn=0\lim_{n \to \infty} x_n = 0.

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