60th Putnam 1999 Problem B3 Let R be the reals. Define f : [0, 1) x [0, 1) → R by f(x, y) = ∑ x m y n , where the sum is taken over all pairs of positive integers (m, n) satisfying m ≥ n/2, n ≥ m/2. Find lim (x, y)→(1, 1) (1 - xy 2 )(1 - x 2 y)f(x, y).
The straight slog approach is to sum the series. This is not too bad. If m = 2r, the y terms are y r (1 + y + ... + y 3r ) = y r (1 - y 3r+1 )/(1 - y). Now summing over r we have ∑ x 2r y r /(1 - y) - ∑ x 2r y 4r+1 /(1 - y) = x 2 y/( (1 - x 2 y)(1 - y) ) - x 2 y 5 /( (1 - x 2 y 4 )(1 - y) ). We deal with odd m in the same way. Let m = 2r - 1, then the y terms are y r (1 + y + ... + y 3r-2 ) = y r (1 - y 3r-1 )/(1 - y). So summing over r, we get xy/( (1 - x 2 y)(1 - y) ) - xy 3 /( (1 - x 2 y 4 )(1 - y) ). After some algebra, we get that the complete sum is xy(1 + x + y + xy - x 2 y 2 )/( (1 - x 2 y)(1 - xy 2 ) ). Hence the required limit is 3. A slightly more elegant approach is to note that the sum over pairs (m, n) with m > n/2 is ∑ x 2n+m y n where the sum is over all m, n ≥ 1. That is just x/(1 - x) x 2 y/(1 - x 2 y). Similarly, the sum over pairs (m, n) with n > m/2 is y/(1 - y) xy 2 /(1 - xy 2 ). Putting the three sums together gives x/(1 - x) y/(1 - y). Hence the required sum is x/(1 - x) y/(1 - y) - x/(1 - x) x 2 y/(1 - x 2 y) - y/(1 - y) xy 2 /(1 - xy 2 ). Unfortunately, things deteriorate from this point on. Getting the expression into the required form xy(1 + x + y + xy - x 2 y 2 )/( (1 - x 2 y)(1 - xy 2 ) ) requires, if anything, rather more slogging than the straight slog approach. 60th Putnam 1999 © John Scholes jscholes@kalva.demon.co.uk 14 Dec 1999