Combinatorics and limit problem

Question

For every natural number $n$ let us consider $a_n$ the greatest natural non-zero number such that: $$\binom {a_n}{n-1} \gt \binom {a_n-1}{n}.$$ Compute $$\lim_{n \to \infty} \frac {a_n}{n}.$$ I started by using the formula for the binomial coefficient, I obtained a second degree inequation in $a_n$, but I can't find the greatest $a_n$. That's where I got stuck. The equation I got is $a_n^2+a_n(1-3n)+n^2-n<0$.

Answer 1

$$ \begin{align} &\, \binom{a_{\small n}-1}{n}\lt\binom{a_{\small n}}{n-1} \,\Rightarrow \frac{(a_{\small n}-1)!}{(n)!\,(a_{\small n}-n-1)!}\lt\frac{(a_{\small n})!}{(n-1)!\,(a_{\small n}-n+1)!} \\[4mm] &\, \Rightarrow 1\lt\frac{a_{\small n}\,n}{(a_{\small n}-n)\,(a_{\small n}-n+1)} \,\Rightarrow\, a_{\small n}^2-(3n-1)a_{\small n}+(n^2-n)\lt0 \\[4mm] &\, \qquad \left\{{\small\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{(3n-1)\pm\sqrt{(3n-1)^2-4(n^2-n)}}{2}=\frac{(3n-1)\pm\sqrt{5n^2-2n+1}}{2}}\right\} \\[4mm] &\, \Rightarrow \left(a_{\small n}-\frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\right)\left(a_{\small n}-\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2}\right)\lt0 \\[4mm] &\, \Rightarrow \frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\lt a_{\small n} \lt\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2} \\[4mm] &\, \Rightarrow a_{\small n}=\color{red}{\left\lfloor\,{\small \frac{(3n-1)+\sqrt{5n^2-2n+1}}{2}}\,\right\rfloor} \\[4mm] &\, \Rightarrow \lim_{n\to\infty}\frac{a_{\small n}}{n}=\lim_{n\to\infty}\frac{(3n-1)+\sqrt{5n^2-2n+1}}{2n} \\[2mm] &\, \quad\qquad\qquad =\lim_{n\to\infty}\frac{3-(1/n)+\sqrt{5-(2/n)+(1/n^2)}}{2}=\color{red}{\frac{3+\sqrt{5}}{2}} \end{align} $$

Answer 2

In fact, applying the definition of the binomial and symplifying a little bit you arrive at the inequality $ a_n^2+(1-3n)a_n+n^2-n<0 $ this has two solutions (for $n$ fixed) :

$\frac{(3n-1)\pm 2\sqrt{5n^2-2n+1}}{4}$

So the largest is

$\frac{(3n-1)+2\sqrt{5n^2-2n+1}}{4}$

and taking the limit

$ \lim_{n\to\infty} \frac{a_n}{n}=\frac{3+2\sqrt 5}{4} $