Problem 1

Let $n$ be a positive integer. Denote by $(M(n))$ the largest integer $m$ satisfying the inequality $$\binom{m}{n-1}>\binom{m-1}{n}.$$ Evaluate the limit $$\lim_{n\rightarrow\infty}\frac{(M(n))}{n}.$$

Solution: $$ \binom{m}{n-1}>\binom{m-1}{n} \iff \frac{m!}{(n-1)!\,(m-n+1)!} > \frac{(m-1)!}{n!\,(m-n-1)!} $$ $$ \iff \frac{m}{(m-n+1)(m-n)}>\frac{1}{n} $$ $$ \iff mn > (m-n)(m-n+1) $$ $$ mn > m^2-2mn+m+n^2-n $$ $$ \iff m^2-3mn+n^2+m-n<0 $$ Rewrite as a quadratic inequality in $m$: $$ m^2+(1-3n)m+n^2-n<0 $$ The discriminant is $$ \Delta=(1-3n)^2-4(n^2-n)=1+9n^2-6n-4n^2+4n=5n^2-2n+1. $$ The two roots are $$ m_1=\frac{(3n-1)-\sqrt{5n^2-2n+1}}{2},\quad m_2=\frac{(3n-1)+\sqrt{5n^2-2n+1}}{2}. $$ $$m_2-m_1=\sqrt{5n^2-2n+1},\quad n\in\mathbb{N}^+.$$ Thus $m_2-m_1>1$, so there exists positive integer solutions for $m$.

The inequality holds for $m_1<m<m_2$. Since $(M(n))$ is the maximum integer satisfying the condition, we have $$ m_2-1\le (M(n))\le m_2. $$ Divide by $n$: $$ \frac{m_2-1}{n}\le\frac{(M(n))}{n}\le\frac{m_2}{n}. $$ Compute the limits $$ \lim_{n\to\infty}\frac{m_2}{n}=\lim_{n\to\infty}\frac{3n-1+\sqrt{5n^2-2n+1}}{2n} =\frac{3}{2}+\frac{\sqrt{5}}{2}, $$ $$ \lim_{n\to\infty}\frac{m_2-1}{n}=\frac{3}{2}+\frac{\sqrt{5}}{2}. $$ By the squeeze theorem, $$ \boxed{\lim_{n\rightarrow\infty}\frac{(M(n))}{n}=\frac{3+\sqrt{5}}{2}}. $$

Problem 2

Let $A=\{(x,y):0\le x,y<1\}$. For $(x,y)\in A$, define $$S(x,y)=\sum_{\substack{m,n\in\mathbb{N}_+\\ \frac12\le\frac{m}{n}\le2}}x^m y^n.$$ Evaluate the limit $$\lim_{\substack{(x,y)\rightarrow(1,1)\\(x,y)\in A}}(1-xy^2)(1-x^2y)S(x,y).$$

Solution: The condition $\dfrac12\le\dfrac{m}{n}\le2$ is equivalent to $\dfrac{n}{2}\le m\le 2n$. We subtract the invalid pairs from all positive integer pairs: $$ S(x,y)=\sum_{m,n\in\mathbb{N}_+}x^m y^n -\sum_{\substack{n\in\mathbb{N}_+\\m\ge 2n+1}} x^m y^n -\sum_{\substack{m\in\mathbb{N}_+\\n\ge 2m+1}} x^m y^n. $$

First term: $$ \sum_{m,n\in\mathbb{N}_+}x^m y^n =\left(\sum_{m=1}^\infty x^m\right)\left(\sum_{n=1}^\infty y^n\right) =\frac{xy}{(1-x)(1-y)}. $$

Second series: $$ \sum_{n=1}^\infty y^n\sum_{m=2n+1}^\infty x^m =\sum_{n=1}^\infty y^n\cdot\frac{x^{2n+1}}{1-x} =\frac{x}{1-x}\sum_{n=1}^\infty (x^2 y)^n =\frac{x}{1-x}\cdot\frac{x^2 y}{1-x^2 y} =\frac{x^3 y}{(1-x)(1-x^2 y)}. $$

Third series: $$ \sum_{m=1}^\infty x^m\sum_{n=2m+1}^\infty y^n =\sum_{m=1}^\infty x^m\cdot\frac{y^{2m+1}}{1-y} =\frac{y}{1-y}\sum_{m=1}^\infty (x y^2)^m =\frac{y}{1-y}\cdot\frac{x y^2}{1-x y^2} =\frac{x y^3}{(1-y)(1-x y^2)}. $$

Hence $$ S(x,y)=\frac{xy}{(1-x)(1-y)}-\frac{x^3 y}{(1-x)(1-x^2y)}-\frac{x y^3}{(1-y)(1-x y^2)}. $$ Combine fractions to obtain: $$ S(x,y)=\frac{xy(1+x+y+xy-x^2y^2)}{(1-xy^2)(1-x^2y)}. $$ Multiply both sides by $(1-xy^2)(1-x^2y)$: $$ (1-xy^2)(1-x^2y)S(x,y)=xy(1+x+y+xy-x^2y^2). $$ Take the limit as $(x,y)\to(1,1)$: $$ \lim_{(x,y)\to(1,1)}xy(1+x+y+xy-x^2y^2)=1\cdot(1+1+1+1-1)=3. $$ Therefore $$ \boxed{\lim_{\substack{(x,y)\rightarrow(1,1)\\(x,y)\in A}}(1-xy^2)(1-x^2y)S(x,y)=3}. $$