If Combinatorics goes Infinity

$\dbinom {m}{n-1}>\dbinom {m-1}{n}$

$\dbinom {m}{n-1}=\cfrac{m!}{(n-1)!(m-n+1)!}$

$\dbinom {m-1}{n}=\cfrac{(m-1)!}{n!(m-n-1)!}$

$\cfrac{m!}{(n-1)!(m-n+1)!}>\cfrac{(m-1)!}{n!(m-n-1)!}$

$\cfrac{m(m-1)!}{(n-1)!(m-n+1)(m-n)(m-n-1)!}>\cfrac{(m-1)!}{n(n-1)!(m-n-1)!}$

$\cfrac{m}{(m-n+1)(m-n)}>\cfrac{1}{n}$

$mn>(m-n+1)(m-n)$

$\cfrac{(m-n+1)(m-n)}{mn}<1$

$\cfrac{m-n+1}{n}\cdot\cfrac{m-n}{m}<1$

$(\cfrac{m}{n}-1+\cfrac{1}{n})(1-\cfrac{n}{m})<1$

$\displaystyle{\lim_{n\to \infty}}(\cfrac{m}{n}-1+\cfrac{1}{n})(1-\cfrac{n}{m})<\displaystyle{\lim_{n\to \infty}}1$

$(\displaystyle{\lim_{n\to \infty}}\cfrac{m}{n}-1+0)(1-\displaystyle{\lim_{n\to \infty}}\cfrac{n}{m})<\displaystyle{\lim_{n\to \infty}}1$

$\displaystyle{\lim_{n\to \infty}}(\cfrac{m}{n}-1)(1-\cfrac{n}{m})<\displaystyle{\lim_{n\to \infty}}1$

$\displaystyle{\lim_{n\to \infty}}(m-n)(m-n)<\displaystyle{\lim_{n\to \infty}}(mn)$

$\displaystyle{\lim_{n\to \infty}}[(m-n)^2-mn]<0$

$\displaystyle{\lim_{n\to \infty}}(m^2-3mn+n^2)<0$

When $m^2-3mn+n^2=0$ ,

$m=\cfrac{3\pm\sqrt{9-4}}{2}n=\cfrac{3\pm\sqrt{5}}{2}n$

$\cfrac{m}{n}=\cfrac{3\pm\sqrt{5}}{2}$

$\therefore \cfrac{3-\sqrt{5}}{2}<\displaystyle{\lim_{n\to \infty}}\cfrac{m}{n}<\cfrac{3+\sqrt{5}}{2}$

We'll focus on $\displaystyle{\lim_{n\to \infty}}\cfrac{m}{n}<\cfrac{3+\sqrt{5}}{2}$

$\displaystyle{\lim_{n\to \infty}}(\cfrac{m}{n}-\cfrac{3+\sqrt{5}}{2})<0=\displaystyle{\lim_{n\to \infty}}0$

$\therefore \displaystyle{\lim_{n\to \infty}}(\cfrac{m_{max}}{n}-\cfrac{3+\sqrt{5}}{2})=0$

$\Rightarrow \displaystyle{\lim_{n\to \infty}}\cfrac{m_{max}}{n}=\boxed{\cfrac{3+\sqrt{5}}{2}}$

I find it a bit hard to explain why the equality holds instead of the inequality in my last part of solution.

This question is taken from William Lowell Putnam Mathematical Competition 2016. Modified.

$\begin{aligned}\frac{m!}{(m-n+1)!(n-1)!}&>\frac{(m-1)!}{(m-n-1)!n!}\\\frac{m}{(m-n+1)(m-n)}&>\frac1n\\m^2-3mn+n^2+m-n&<0\end{aligned}$ Try to write $m$ in terms of $n$ , it can help us to calculate the limit. Hence, $m^2-(3n-1)m+n^2-n<0$ $\frac{3n-1-\sqrt{5n^2-2n+1}}{2}<m<\frac{3n-1+\sqrt{5n^2-2n+1}}{2}$ Initially, I thought I have to consider how to make $m$ an integer, but then I knew that when $n$ goes infinity, the value of $m$ will approach the maximum value $\dfrac{3n-1+\sqrt{5n^2-2n+1}}{2}$ , so the maximum of $m$ when $n$ goes infinity will be $\lim_{n\to\infty}m_{max}\sim\lim_{n\to\infty}\dfrac{3n-1+\sqrt{5n^2-2n+1}}{2}$ Divided by $n$ , $\begin{aligned}\lim_{n\to\infty}\frac mn&=\lim_{n\to\infty}\frac{3-\dfrac1n+\sqrt{5-\dfrac2n+\dfrac1{n^2}}}{2}\\&=\frac{3+\sqrt5}{2}\end{aligned}$ So the answer will be $3+5+2=\boxed{10}$ .

Additional Insight: The final value is $2.618$ which is the golden ratio added 1! Wow!