HMMT 二月 2006 · 冲刺赛 · 第 35 题

HMMT February 2006 — Guts Round — Problem 35

[12] A sequence is defined by A = 0 , A = 1 , A = 2, and, for integers n ≥ 3, 0 1 2 A + A + A 1 n − 1 n − 2 n − 3 A = + n 4 2 3 n − n Compute lim A . N →∞ N

解析

A sequence is defined by A = 0 , A = 1 , A = 2, and, for integers n ≥ 3, 0 1 2 A + A + A 1 n − 1 n − 2 n − 3 A = + n 4 2 3 n − n Compute lim A . N →∞ N 2 13 π Answer: − . 6 12 Solution: If we sum the given equation for n = 3 , 4 , 5 , . . . , N , we obtain N N ∑ ∑ A + A + A 1 n − 1 n − 2 n − 3 A = + n 4 2 3 n − n n =3 n =3 This reduces dramatically to N ∑ 2 A A 2 A A 1 N − 1 N − 2 1 0 A + + = A + + + () N 2 4 2 3 3 3 3 n − n n =3 Let lim A = L . Under this limit, the left hand side of () is simply 2 L . We N →∞ N compute the sum on the right with the help of partial fractions N ∞ ∑ ∑ 1 1 1 lim = − 4 2 2 2 N →∞ n − n n − 1 n n =3 n =3 ( ) ( ) ∞ ∞ ∑ ∑ 1 1 1 1 1 1 = − + + − 2 2 2 2 n − 1 n + 1 1 2 n n =3 n =1 ( ) 2 1 1 1 5 π = + + − 2 2 3 4 6 2 5 π = − 3 6 ( ) 2 2 1 2 1 5 π 13 π With this we easily find L = · 2 + · 1 + · 0 + − = − , and we are done. 2 3 3 3 6 6 12