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).

**Solution**

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.

© John Scholes

jscholes@kalva.demon.co.uk

14 Dec 1999