### 60th Putnam 1999

Problem B3

Let R be the reals. Define f :[0, 1) x [0, 1) → R by f(x, y) = ∑ xmyn, 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 - xy2)(1 - x2y)f(x, y).

Solution

The straight slog approach is to sum the series. This is not too bad. If m = 2r, the y terms are yr(1 + y + ... + y3r) = yr(1 - y3r+1)/(1 - y). Now summing over r we have ∑ x2ryr/(1 - y) - ∑ x2ry4r+1/(1 - y) = x2y/( (1 - x2y)(1 - y) ) - x2y5/( (1 - x2y4)(1 - y) ). We deal with odd m in the same way. Let m = 2r - 1, then the y terms are yr(1 + y + ... + y3r-2) = yr(1 - y3r-1)/(1 - y). So summing over r, we get xy/( (1 - x2y)(1 - y) ) - xy3/( (1 - x2y4)(1 - y) ). After some algebra, we get that the complete sum is xy(1 + x + y + xy - x2y2)/( (1 - x2y)(1 - xy2) ). 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 ∑ x2n+myn where the sum is over all m, n ≥ 1. That is just x/(1 - x) x2y/(1 - x2y). Similarly, the sum over pairs (m, n) with n > m/2 is y/(1 - y) xy2/(1 - xy2). 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) x2y/(1 - x2y) - y/(1 - y) xy2/(1 - xy2). Unfortunately, things deteriorate from this point on. Getting the expression into the required form xy(1 + x + y + xy - x2y2)/( (1 - x2y)(1 - xy2) ) requires, if anything, rather more slogging than the straight slog approach.