HMMT 二月 2006 · 冲刺赛 · 第 37 题

HMMT February 2006 — Guts Round — Problem 37

解析

Compute ∞ ∑ 2 n + 5 n 3 2 2 · ( n + 7 n + 14 n + 8) n =1 137 Answer: − 8 ln 2 24 Solution: First, we manipulate using partial fractions and telescoping: ( ) ∞ ∞ ∑ ∑ 2 n + 5 1 1 2 1 1 = · − − n 3 2 n 2 · ( n + 7 n + 14 n + 8) 2 2 n + 1 n + 2 n + 4 n =1 n =1 ∞ ∑ 1 1 1 = − n 4 2 2 · ( n + 4) n =1 ∑ n ∞ r Now, consider the function f ( r, k ) := . We have k n =1 n [ ] ∞ ∞ ∞ n n n − 1 ∑ ∑ ∑ ∂f ( r, k ) ∂ r ∂ r r 1 = = = = f ( r, k − 1) k k k − 1 ∂r ∂r n ∂r n n r n =1 n =1 n =1 ∞ n ∑ df ( r, 1) 1 r 1 r 1 = = · = 0 dr r n r 1 − r 1 − r n =1 ∫ dr f ( r, 1) = = − ln(1 − r ) + f (0 , 1) 1 − r ( ) ∑ ∞ 1 1 By inspection, f (0 , 1) = 0, so f , 1 = = ln(2). It is easy to compute the n n =1 2 n · 2 ( ) ∑ ∞ 1 1 131 desired sum in terms of f , 1 , and we find = 16 ln(2) − . Hence, our n n =1 2 2 ( n +4) 12 137 final answer is − 8 ln(2). 24