Russian 1999

Let {x} denote the fractional part of x. Show that {√1} + {√2} + {√3} + ... + {√(n²)} ≤ (n² - 1)/2.

Solution

For i = 1, 2, ... , 2n we have {√(n²+i) } = √(n²+i) - n < (n+i/2n) - n = i/2n. So {√(n²+1) } + {√(n²+2) } + ... + {√(n²+2n+1) } = {√(n²+1) } + {√(n²+2) } + ... + {√(n²+2n) } < (1 + 2 + ... + 2n)/2n = (2n+1)/2. Hence {√1} + {√2} + {√3} + ... + {√(n²)} ≤ 3/2 + 5/2 + ... + (2n-1)/2 = (n-1)(2n+2)/2.

Russian 1999

© John Scholes
jscholes@kalva.demon.co.uk
4 March 2004
Last corrected/updated 4 Mar 04