HMMT 二月 2025 · ALGNT 赛 · 第 8 题

HMMT February 2025 — ALGNT Round — Problem 8

Define sgn( x ) to be 1 when x is positive, − 1 when x is negative, and 0 when x is 0. Compute ∞ n X sgn(sin(2 )) . n 2 n =1 (The arguments to sin are in radians.)

解析

Define sgn( x ) to be 1 when x is positive, − 1 when x is negative, and 0 when x is 0. Compute ∞ n X sgn(sin(2 )) . n 2 n =1 (The arguments to sin are in radians.) Proposed by: Karthik Venkata Vedula 2 Answer: 1 − π Solution: Note that each of following is equivalent to the next. n • sgn(sin(2 )) = +1. n • 0 < 2 mod 2 π < π . n 2 • 0 < mod 2 < 1. π 1 • The n th digit after the decimal point in the binary representation of is 0. π n Similarly, sgn(sin(2 )) = − 1 if and only if the n -th digit after the decimal point in the binary repre- 1 n sentation of is 1. In particular, if a is the n -th digit, then sgn(sin(2 )) = 1 − 2 a . Thus, the desired n n π sum is ! ! ∞ ∞ ∞ ∞ n X X X X sgn(sin(2 )) 1 − 2 a 1 a 2 n n = = − 2 = 1 − . n n n n 2 2 2 2 π n =1 n =1 n =1 n =1