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Algebra Difficulty 4.9 AIME Prove it Turkey

Let $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ be sequences of real numbers such that $a_0 = 40$ , $b_0 = 41$ and for all $n \ge 0$ we have
$a_{n+1} = a_n + \frac{1}{b_n} \quad \text{and} \quad b_{n+1} = b_n + \frac{1}{a_n}$
Find the smallest positive integer $k$ such that $a_k > 80$ .

Solution

Answer: 2460.
Since
$\frac{a_{n+1}}{b_{n+1}} = \frac{a_n + \frac{1}{b_n}}{b_n + \frac{1}{a_n}} = \frac{a_n}{b_n}$
we get that $\frac{a_n}{b_n}$ is a constant and hence equals $\frac{40}{41}$ . Therefore, $a_k > 80$ is equivalent to $b_k > 82$ and $a_k b_k > 80 \cdot 82 = 6560$ . Multiplying both sequences we get
$a_{n+1} b_{n+1} = a_n b_n + 2 + \frac{1}{a_n b_n}$
Hence, for any $k$ we have $a_k b_k > 40 \cdot 41 + 2k$ which means $a_k b_k > 6560$ for $k = 2460$ . Moreover, if $k = 2459$ , we have
$a_k b_k = 6560 - 2 + \sum_{i=0}^{k-1} \frac{1}{a_i b_i} < 6560 - 2 + \frac{2459}{1640} < 6560.$
Thus, the minimal value of $k$ satisfying conditions is 2460.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.