Sequence and inequality

Algebra Level 3

{ a n } \{a_n\} is a sequence such that a 1 = 1 a_1=1 , a n + 1 − a n = ( − 1 2 ) n a_{n+1}-a_{n}=\left(-\dfrac{1}{2}\right)^n for positive integer n ≥ 1 n \geq 1 .

If there exists a positive integer n n which satisfies ( a n − λ ) ( a n + 1 − λ ) < 0 (a_{n}-\lambda)(a_{n+1}-\lambda)<0 for real λ \lambda , what is the range of λ \lambda ?

( 2 3 , 1 ) (\dfrac{2}{3},1) ( 2 3 , 5 6 ) (\dfrac{2}{3},\dfrac{5}{6}) ( 1 2 , 2 ) (\dfrac{1}{2},2) ( 1 2 , 1 ) (\dfrac{1}{2},1)

Alice Smith by Alice Smith , 20, China

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1 solution

Patrick Corn
Sep 9, 2019

It's not hard to solve explicitly for a n , a_n, but it's not necessary either. It's clear from the description of the a n a_n that a 2 < a 4 < a 6 < ⋯ < a 5 < a 3 < a 1 , a_2 < a_4 < a_6 < \cdots < a_5 < a_3 < a_1, and the condition ( a n − λ ) ( a n + 1 − λ ) < 0 (a_n - \lambda)(a_{n+1}-\lambda) < 0 is equivalent to saying that λ \lambda is between a n a_n and a n + 1 . a_{n+1}. So if λ \lambda satisfies the condition for some n , n, it must satisfy it for n = 1. n=1. That is, λ \lambda is in the range ( a 2 , a 1 ) = ( 1 2 , 1 ) . (a_2, a_1) = (\frac12, 1).

(If you solve explicitly for a n , a_n, you can show that the only λ \lambda that satisfies the condition for all n n is λ = 2 / 3. \lambda = 2/3. )

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