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Difficulty: 7/102021 USAMO 2021 (Q1)

Let be a sequence of positive real numbers such that for all . Prove that the sequence diverges.

Options:

  • A.

    The relation holds only for sufficiently large values in the system.

  • Divergence proven via mathematical induction and positive growth bounds

  • C.

    Divergence proven via mathematical induction and negative growth bounds

  • D.

    No such configuration exists under the given conditions.

Guide / Hint

Hint 1: Use proof by contradiction: assume the sequence converges to a limit .

Hint 2: Take limits of the recurrence relation to show that the only possible limit is .

Hint 3: Show that if , then the terms strictly increase and grow away from 1/2, creating a contradiction with convergence.

Solution

Step 1: Assume for contradiction that the sequence converges to a finite limit . Since the terms are positive real numbers, we must have .

Step 2: Taking the limit on both sides of the inequality :

Step 3: The only real number satisfying this inequality is .

Step 4: Analyze the growth of the terms. If , then . By mathematical induction and looking at the growth difference , we find that the sequence must grow beyond any bound if it ever exceeds , violating the convergence limit. Hence, the sequence must diverge.

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