HMMT 二月 2006 · 冲刺赛 · 第 40 题

HMMT February 2006 — Guts Round — Problem 40

解析

Compute ∞ ∑ 3 k + 1 k +1 · ( − 1) . 3 2 2 k + k k =1 2 π π Answer: + − 2 + ln 2 12 2 Solution: Via partial fraction decomposition we write the sum as ( ) ∞ ∑ 1 2 1 k +1 − + ( − 1) 2 k 1 + 2 k k k =1 Now recall that ∞ 2 ∑ 1 π = = S 1 2 k 6 k =1 ∞ k +1 ∑ ( − 1) = ln(2) = S 2 k k =1 ∞ k +1 ∑ ( − 1) π = = S 3 2 k − 1 4 k =1 14 Manipulating (1), we deduce ( ) ∞ ∞ ∞ k +1 ∑ ∑ ∑ ( − 1) 1 1 = − 2 · 2 2 2 k k (2 k ) k =1 k =1 k =1 2 2 2 π π π = − 2 / 4 · = = S 4 6 6 12 It is then easily seen that the answer is equal to S + 2 · S − 2 + S . 2 3 4