Combinatorics and limit problem 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 $a_n^2+a_n(1-3n)+n^2-n<0$.
 A: $$ 
\begin{align} 
&\, \binom{a_{\small n}-1}{n}\lt\binom{a_{\small n}}{n-1} \,\Rightarrow \frac{(a_{\small n}-1)!}{(n)!\,(a_{\small n}-n-1)!}\lt\frac{(a_{\small n})!}{(n-1)!\,(a_{\small n}-n+1)!} \\[4mm] 
&\, \Rightarrow 1\lt\frac{a_{\small n}\,n}{(a_{\small n}-n)\,(a_{\small n}-n+1)} \,\Rightarrow\, a_{\small n}^2-(3n-1)a_{\small n}+(n^2-n)\lt0 \\[4mm] 
&\, \qquad \left\{{\small\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{(3n-1)\pm\sqrt{(3n-1)^2-4(n^2-n)}}{2}=\frac{(3n-1)\pm\sqrt{5n^2-2n+1}}{2}}\right\} \\[4mm] 
&\, \Rightarrow \left(a_{\small n}-\frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\right)\left(a_{\small n}-\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2}\right)\lt0 \\[4mm] 
&\, \Rightarrow \frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\lt a_{\small n} \lt\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2} \\[4mm] 
&\, \Rightarrow a_{\small n}=\color{red}{\left\lfloor\,{\small \frac{(3n-1)+\sqrt{5n^2-2n+1}}{2}}\,\right\rfloor} \\[4mm] 
&\, \Rightarrow \lim_{n\to\infty}\frac{a_{\small n}}{n}=\lim_{n\to\infty}\frac{(3n-1)+\sqrt{5n^2-2n+1}}{2n} \\[2mm] 
&\, \quad\qquad\qquad =\lim_{n\to\infty}\frac{3-(1/n)+\sqrt{5-(2/n)+(1/n^2)}}{2}=\color{red}{\frac{3+\sqrt{5}}{2}}
\end{align} 
$$ 
A: 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}
$
