HMMT 二月 2015 · 团队赛 · 第 10 题

HMMT February 2015 — Team Round — Problem 10

[ 40 ] The sequences of real numbers { a } and { b } satisfy a = ( a − 1)( b + 1) and i i n +1 n − 1 n i =1 i =1 b = a b − 1 for n ≥ 2, with a = a = 2015 and b = b = 2013. Evaluate, with proof, the n +1 n n − 1 1 2 1 2 infinite sum ( ) ∞ ∑ 1 1 b − . n a a n +1 n +3 n =1

解析

[ 40 ] The sequences of real numbers { a } and { b } satisfy a = ( a − 1)( b + 1) and i i n +1 n − 1 n i =1 i =1 b = a b − 1 for n ≥ 2, with a = a = 2015 and b = b = 2013. Evaluate, with proof, the n +1 n n − 1 1 2 1 2 infinite sum ( ) ∞ ∑ 1 1 b − . n a a n +1 n +3 n =1 1 4058211 Answer: 1 + OR First note that a and b are weakly increasing and tend to n n 2014 · 2015 4058210 infinity. In particular, a , b ∈ { / 0 , − 1 , 1 } for all n . n n For n ≥ 1, we have a = ( a − 1)( b + 1) = ( a − 1)( a b ), so n +3 n +1 n +2 n +1 n +1 n b 1 1 1 n = = − . a a ( a − 1) a − 1 a n +3 n +1 n +1 n +1 n +1 Therefore, ( ) ∞ ∞ ∑ ∑ b b b 1 1 n n n − = − − a a a a − 1 a n +1 n +3 n +1 n +1 n +1 n =1 n =1 ∞ ∑ b + 1 1 n = − . a a − 1 n +1 n +1 n =1 Team a n +1 Furthermore, b + 1 = for n ≥ 2. So the sum over n ≥ 2 is n a − 1 n − 1 ( ) ( ) ∞ N ∑ ∑ 1 1 1 1 − = lim − a − 1 a − 1 N →∞ a − 1 a − 1 n − 1 n +1 n − 1 n +1 n =2 n =2 ( ) 1 1 1 1 = + − lim + N →∞ a − 1 a − 1 a − 1 a − 1 1 2 N N +1 1 1 = + . a − 1 a − 1 1 2 Hence the final answer is ( ) ( ) b + 1 1 1 1 1 − + + . a a − 1 a − 1 a − 1 2 2 1 2 Cancelling the common terms and putting in our starting values, this equals 2014 1 1 1 1