HMMT 二月 2010 · CALC 赛 · 第 10 题

HMMT February 2010 — CALC Round — Problem 10

[ 8 ] Let f ( n ) = . Then there exists constants γ , c , and d such that k k =1 c d 1 f ( n ) = ln( n ) + γ + + + O ( ) , 2 3 n n n 1 1 where the O ( ) means terms of order or lower. Compute the ordered pair ( c, d ). 3 3 n n

解析

[ 8 ] Let f ( n ) = . Then there exists constants γ , c , and d such that k k =1 c d 1 f ( n ) = ln( n ) + γ + + + O ( ) , 2 3 n n n 1 1 where the O ( ) means terms of order or lower. Compute the ordered pair ( c, d ). 3 3 n n 1 1 k 1 Answer: ( , − ) From the given formula, we pull out the term from O ( ), making f ( n ) = 3 4 2 12 n n c d k 1 log( n ) + γ + + + + O ( ). Therefore, 2 3 4 n n n n ( ) ( ) ( ) ( ) ( ) n + 1 1 1 1 1 1 1 1 f ( n +1) − f ( n ) = log − c − − d − − k − + O . 2 2 3 3 4 n n n + 1 n ( n + 1) n ( n + 1) n 1 1 For the left hand side, f ( n + 1) − f ( n ) = . By substituting x = , the formula above becomes n +1 n 2 x 1 x + 2 x + 3 x + 3 2 3 4 4 = log (1 + x ) − cx · − dx · − kx · + O ( x ) . 2 3 x + 1 x + 1 ( x + 1) ( x + 1) 1 1 Because x is on the order of , is on the order of a constant. Therefore, all the terms in the 3 n ( x +1) 2 x +3 x +3 4 4 4 expansion of kx · are of order x or higher, so we can collapse it into O ( x ). Using the Taylor 3 ( x +1) expansions, we get ( ) ( ) ( ) ( ) 1 1 2 4 2 3 2 3 4 x 1 − x + x + O x = x − x + x − cx (1 − x ) − dx (2) + O x . 2 3 1 1 Coefficient comparison gives c = and d = − . 2 12 Calculus Subject Test