HMMT 二月 2005 · 冲刺赛 · 第 22 题

HMMT February 2005 — Guts Round — Problem 22

[9] Find 2 4 8 { ln(1 + e ) } + { ln(1 + e ) } + { ln(1 + e ) } + { ln(1 + e ) } + · · · , where { x } = x − ⌊ x ⌋ denotes the fractional part of x .

解析

Find 2 4 8 { ln(1 + e ) } + { ln(1 + e ) } + { ln(1 + e ) } + { ln(1 + e ) } + · · · , where { x } = x − b x c denotes the fractional part of x . Solution: 1 − ln( e − 1) k k k 2 k 2 2 Since ln(1 + e ) is just larger than 2 , its fractional part is ln(1 + e ) − ln e = k − 2 ln(1 + e ). But now notice that n ∏ k n +1 2 2 2 − 1 (1 + x ) = 1 + x + x + · · · + x . k =0 (This is easily proven by induction or by noting that every nonnegative integer less n +1 than 2 has a unique ( n +1)-bit binary expansion.) If | x | < 1, this product converges 1 to as n goes to infinity. Therefore, 1 − x ∞ ∞ ∑ ∏ 1 e k k − 2 − 1 2 ln(1 + e ) = ln (1 + ( e ) ) = ln = ln = 1 − ln( e − 1) . − 1 1 − e e − 1 k =0 k =0