24th IMO 1983 shortlist

Find the greatest integer not exceeding 1 + 1/2^k + 1/3^k + ... + 1/N^k, where k = 1982/1983 and N = 2¹⁹⁸³.

Solution

We have (a¹⁹⁸³ - b¹⁹⁸³) = (a - b)(a¹⁹⁸² + a¹⁹⁸¹b + a¹⁹⁸⁰b² + ... + b¹⁹⁸²). So if a > b, we have (a¹⁹⁸³ - b¹⁹⁸³) < (a - b) 1983 a¹⁹⁸². Take a = m^1/1983, b = (m-1)^1/1983 and this becomes (1/m)^1982/1983 < 1983 (m^1/1983 - (m-1)^1/1983). Summing we get 1/2^k + 1/3^k + ... + 1/N^k < 1983( N^1/1983 - 1) = 1983 (2 - 1). So 1 + 1/2^k + 1/3^k + ... + 1/N^k < 1984.

Similarly, (a¹⁹⁸³ - b¹⁹⁸³) > (a - b) 1983 b¹⁹⁸², so (1/(m-1))^k > 1983 (m^1/1983 - (m-1)^1/1983). Summing gives 1 + 1/2^k + 1/3^k + ... + 1/(N-1)^k > 1983(2 - 1) = 1983. Hence 1 + 1/2^k + 1/3^k + ... + 1/N^k lies strictly between 1983 and 1984. So the integer part is 1983.

24th IMO shortlist 1983

© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 2003
Last corrected/updated 26 Nov 03