Let be a sequence of positive real numbers such that for all . Prove that the sequence diverges.
The relation holds only for sufficiently large values in the system.
Divergence proven via mathematical induction and positive growth bounds
Divergence proven via mathematical induction and negative growth bounds
No such configuration exists under the given conditions.
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.
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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