Two infinite sequences { a n } and { b n } satisfy the following recurrence relations: ⎩ ⎪ ⎨ ⎪ ⎧ a n + 1 = a n + 2 b n b n + 1 = a n + b n . Given that a n b n > 0 for any positive integer n , n → ∞ lim a n b n = D N Q , where positive integers D and N are coprime and Q is a square-free integer.
Find the value of N + Q + D .
This problem is a part of <Calculus - Inadequate Information> series .
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a n and b n never converges, so it is natural to try and find the limiting value of a n b n .
Divide the second equation by the first.
a n + 1 b n + 1 = a n + 2 b n a n + b n . Divide the numerator and denominator by a n on the right side.
a n + 1 b n + 1 = 1 + 2 ⋅ a n b n 1 + a n b n . Set a n b n = c n .
c n + 1 = 1 + 2 c n 1 + c n . It is clear that c n > 0 for every natural number n , and therefore { c n } converges.
Let n → ∞ lim c n = α and we see that
α = 1 + 2 α 1 + α , or 2 α 2 = 1 .
Since α > 0 , we see that n → ∞ lim a n b n = n → ∞ lim c n = α = 2 2 .
Therefore N + Q + D = 1 + 2 + 2 = 5 .