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Recurrence sequence grows like sqrt(2log n)

Source: VJIMC 2024, Category II, Problem 3

April 14, 2024
recurrence relationSequenceslimitasymptotic behaviourreal analysis

Problem Statement

Let a1>0a_1>0 and for n≥1n \ge 1 define an+1=an+1a1+a2+⋯+an.a_{n+1}=a_n+\frac{1}{a_1+a_2+\dots+a_n}. Prove that lim⁡n→∞an2ln⁡n=2.\lim_{n \to \infty} \frac{a_n^2}{\ln n}=2.
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