Recurrence with a square root

Task number: 3876

For the sequence given by the recurrence \(a_0=2, a_1=8, a_{n+2}=\sqrt{a_n a_{n+1}}\) determine \(\lim_{n \to \infty} a_n\).

  • Solution

    We introduce a sequence \(b_n=\log_2 a_n\).

    It holds that \(b_0=1\), \(b_1=3\) and \(b_{n+2}=\frac{b_{n+1}+b_n}2\).

    According to the previous task, \(\lim_{n\to\infty} b_n=\frac{b_0+2b_2}3=\frac73\).

  • Answer

    The limit is \(2^{7/3}=4 {\cdot} 2^{1/3}\).

Difficulty level: Easy task (using definitions and simple reasoning)
Routine calculation training
Cs translation
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