Sequence gets beyond 2018

Algebra Level 3

Let a n a_n and b n b_n be sequences of positive reals such that, a n + 1 = a n + 1 2 b n a_{n+1}= a_n + \dfrac{1}{2b_n} and b n + 1 = b n + 1 2 a n b_{n+1}= b_n + \dfrac{1}{2a_n} for all n ∈ N n\in\mathbb N . Then max ( a 2018 , b 2018 ) > \text{max}\left(a_{2018}, b_{2018}\right) > k . (k should be the maximum possible value).


The answer is 44.

Saikat Sengupta by Saikat Sengupta , 18, India

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1 solution

Saikat Sengupta
Nov 30, 2018

Multiplying the recurrence, we have a n + 1 b n + 1 = 1 + a n b n + 1 4 a n b n ≥ 2 a_{n+1}b_{n+1}=1+a_nb_n+ \frac{1}{4a_nb_n} \ge 2 . This implies that a n b n > n + 1 a_nb_n > n+1 for all n ∈ N n \in \mathbb{N} so max ( a 2018 , b 2018 ) > 2019 > 44 \text{max}\left(a_{2018}, b_{2018}\right) >\sqrt{2019} > 44 .

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