PUMaC 2020 · 个人决赛（A 组） · 第 1 题

PUMaC 2020 — Individual Finals (Division A) — Problem 1

Let a , . . . , a be a sequence of real numbers such that a = 2 , and a a = a − a . 1 2020 1 n n n − 1 n − 1 1 Prove that a < . 2020 2019 2 − 1

解析

Let a , . . . , a be a sequence of real numbers such that a = 2 , and a a = a − a . 1 2020 1 n n n − 1 n − 1 1 Prove that a < . 2020 2019 2 − 1 − 2020+ i Proof. We will prove by induction that 0 < a ≤ 2 for i = 1 , . . . , 2020. The base follows i from the definition of a . 1 a i Suppose the statement holds for i . Then a = , from the recurrence equation. By the i +1 2 1 − a i a 1 i − 2020+ i inductive hypothesis, first we see that a > 0. Furthermore, ≤ 2 . It is i +1 2 2 1 − a 1 − a i i 1 − 2020+ i enough to prove that < 2 which follows from a < 2 . i 2 1 − a i ∑ 1 1 1 1 2019 From the recurrence equation, we get that a = − . Then − = a ≤ n − 1 i i =1 a a a a n − 1 n 1 2020 ∑ 1 1 2019 − 2020+ i 2019 2 < 1. Then 2 − < 1, from which it follows that a < . 2020 i =1 2019 a 2 − 1 2020 Remark: This technique can be repeated to get an even better estimate by using a similar ∑ 1 2019 2019 estimate for a < , yielding a < . i i i =1 2019 2019 2 − 1 2 − 1 Proposed by Aleksa Milojevi´ c and Igor Medvedev.