MHB What is the limit of a special sum at the point (1,1)?

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The discussion centers on evaluating the limit of a special sum \( S(x,y) \) as \( (x,y) \) approaches \( (1,1) \) within the defined set \( A \). The sum \( S(x,y) \) is defined over pairs of positive integers \( (m,n) \) that satisfy \( \frac{1}{2} \leq \frac{m}{n} \leq 2 \). The limit expression involves the factors \( (1-xy^2)(1-x^2y) \) multiplied by \( S(x,y) \). No participants provided answers to the Problem of the Week, but the solution is credited to Kiran Kedlaya and his team. The problem is noted as part of the 1999 William Lowell Putnam Mathematical Competition.
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Here is this week's POTW:

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Let $A=\{(x,y):0\leq x,y<1\}$. For $(x,y)\in A$, let
\[S(x,y) = \sum_{\frac{1}{2}\leq \frac{m}{n}\leq 2} x^m y^n,\]
where the sum ranges over all pairs $(m,n)$ of positive integers satisfying the indicated inequalities. Evaluate
\[\lim_{(x,y)\rightarrow (1,1), (x,y)\in A} (1-xy^2)(1-x^2y)S(x,y).\]

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Re: Problem Of The Week # 256 - Mar 18, 2017

This was Problem B-3 in the 1999 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

We first note that
\[
\sum_{m,n > 0} x^m y^n = \frac{xy}{(1-x)(1-y)}.
\]
Subtracting $S$ from this gives two sums, one of which is
\[
\sum_{m \geq 2n+1} x^m y^n = \sum_n y^n \frac{x^{2n+1}}{1-x}
= \frac{x^3y}{(1-x)(1-x^2y)}
\]
and the other of which sums to $xy^3/[(1-y)(1-xy^2)]$. Therefore
\begin{align*}
S(x,y) &= \frac{xy}{(1-x)(1-y)} - \frac{x^3y}{(1-x)(1-x^2y)} \\
&\qquad - \frac{xy^3}{(1-y)(1-xy^2)} \\
&= \frac{xy(1+x+y+xy-x^2y^2)}{(1-x^2y)(1-xy^2)}
\end{align*}
and the desired limit is
\[
\lim_{(x,y) \to (1,1)} xy(1+x+y+xy-x^2y^2) = 3.
\]