PUMaC 2017 · 个人决赛（A 组） · 第 2 题

PUMaC 2017 — Individual Finals (Division A) — Problem 2

Let a , a , a , . . . be a monotonically decreasing sequence of positive real numbers converging to 1 2 3 ∞ ∞ ∑ ∑ 2017 a 2 i zero. Suppose that diverges. Show that a also diverges. You may assume in your i i i =1 i =1 ∞ ∞ ∑ ∑ 1 proof that converges for all real numbers p > 1. (A sum b of positive real numbers b p i i i i =1 i =1 diverges if for each real number N there is a positive integer k such that b + b + · · · + b > N .) 1 2 k

解析

Fix any ε > 0 such that 1 − (2 − 1) ε > 0. First, we will show by induction on k that k ∞ 2 ∑ a i k 1 − (2 − 1) ε i i =1 diverges for each k ∈ { 0 , 1 , 2 , . . . , 2017 } . For the base case, when k = 0, k ∞ ∞ 2 ∑ ∑ a a i i = , k 1 − (2 − 1) ε i i i =1 i =1 which diverges by hypothesis. For the inductive step, suppose k ∞ 2 ∑ a i k 1 − (2 − 1) ε i i =1 diverges. Then, ( ) ( ) ( ) 2 k +1 k N N N 2 2 ∑ ∑ ∑ a 1 a i i · ≥ k +1 k 1+ ε 1 − (2 − 1) ε 1 − (2 − 1) ε i i i i =1 i =1 i =1 1 ∞ ∑ 1 by the Cauchy-Schwarz inequality. Since converges, and the RHS diverges, 1+ ε i i =1 k +1 N 2 ∑ a i k +1 1 − (2 − 1) ε i i =1 diverges, so our induction is complete. 2017 Finally, since 1 − (2 − 1) ε > 0, 2017 N N 2 ∑ ∑ 2017 a 2 i a ≥ , 2017 i 1 − (2 − 1) ε i i =1 i =1 ∞ ∑ 2017 2 so a diverges as well, as desired. i i =1 Problem written by Mel Shu