Calculus - Inadequate Information #1

Calculus Level 3

Two infinite sequences { a n } \{a_n\} and { b n } \{b_n\} satisfy the following recurrence relations: { a n + 1 = a n + 2 b n b n + 1 = a n + b n . \cases{a_{n+1}=a_n+2b_n \\\\ b_{n+1}=a_n+b_n.} Given that a n b n > 0 a_nb_n>0 for any positive integer n , n, lim n b n a n = N Q D , \lim_{n\to\infty}{\dfrac{b_n}{a_n}}=\dfrac{N\sqrt{Q}}{D}, where positive integers D D and N N are coprime and Q Q is a square-free integer.

Find the value of N + Q + D . N+Q+D.


This problem is a part of <Calculus - Inadequate Information> series .


The answer is 5.

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1 solution

Boi (보이)
Aug 23, 2017

a n a_n and b n b_n never converges, so it is natural to try and find the limiting value of b n a n . \dfrac{b_n}{a_n}.

Divide the second equation by the first.

b n + 1 a n + 1 = a n + b n a n + 2 b n . \dfrac{b_{n+1}}{a_{n+1}}=\dfrac{a_n+b_n}{a_n+2b_n}. Divide the numerator and denominator by a n a_n on the right side.

b n + 1 a n + 1 = 1 + b n a n 1 + 2 b n a n . \dfrac{b_{n+1}}{a_{n+1}}=\dfrac{1+\dfrac{b_n}{a_n}}{1+2\cdot\dfrac{b_n}{a_n}}. Set b n a n = c n . \dfrac{b_n}{a_n}=c_n.

c n + 1 = 1 + c n 1 + 2 c n . c_{n+1}=\dfrac{1+c_n}{1+2c_n}. It is clear that c n > 0 c_n>0 for every natural number n , n, and therefore { c n } \{c_n\} converges.

Let lim n c n = α \displaystyle \lim_{n\to\infty} c_n=\alpha and we see that

α = 1 + α 1 + 2 α , \alpha=\dfrac{1+\alpha}{1+2\alpha}, or 2 α 2 = 1. 2\alpha^2=1.

Since α > 0 , \alpha>0, we see that lim n b n a n = lim n c n = α = 2 2 . \displaystyle \lim_{n\to\infty}\dfrac{b_n}{a_n}=\lim_{n\to\infty}c_n=\alpha=\dfrac{\sqrt{2}}{2}.

Therefore N + Q + D = 1 + 2 + 2 = 5 . N+Q+D=1+2+2=\boxed{5}.

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