FM November 2021 p11 q03
4276
The sequence of real numbers \(a_1, a_2, a_3, \ldots\) is such that \(a_1 = 1\) and
\(a_{n+1} = \left( a_n + \frac{1}{a_n} \right)^3.\)
(a) Prove by mathematical induction that \(\ln a_n \geq 3^{n-1} \ln 2\) for all integers \(n \geq 2\).
[You may use the fact that \(\ln \left( x + \frac{1}{x} \right) > \ln x\) for \(x > 0\).]
(b) Show that \(\ln a_{n+1} - \ln a_n > 3^{n-1} \ln 4\) for \(n \geq 2\).
Solution
(a) Base case: For \(n = 2\), \(a_2 = 2^3\) leading to \(\ln a_2 = 3 \ln 2\), so true when \(n = 2\).
Inductive step: Assume \(\ln a_k \geq 3^{k-1} \ln 2\).
Then \(\ln a_{k+1} = 3 \ln \left( a_k + \frac{1}{a_k} \right) > 3 \ln a_k \geq 3^k \ln 2\).
Thus, true for \(n = k+1\). By induction, true for all integers \(n \geq 2\).
(b) \(3 \ln \left( a_n + \frac{1}{a_n} \right) - \ln a_n > 2 \ln a_n\)
\(> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4\)
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