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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 $a_n^2+a_n(1-3n)+n^2-n<0$.

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    $\begingroup$ show your working so far. clearly obtaining the second degree inequation is an essential step. if you have done this, write it down in your question. that is more useful for others, and shows you have engaged with the problem $\endgroup$ Feb 4, 2017 at 12:13

2 Answers 2

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

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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} $

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  • $\begingroup$ I got $a_n^2+a_n(1-3n)+n^2-n<0$ $\endgroup$
    – M. Stefan
    Feb 4, 2017 at 15:21
  • $\begingroup$ @Marczinga can you help me? $\endgroup$
    – M. Stefan
    Feb 4, 2017 at 15:29
  • $\begingroup$ Oh yes sorry! My bad! I will edit my answer to fix it $\endgroup$
    – Maczinga
    Feb 4, 2017 at 15:38
  • $\begingroup$ The formula is wrong again, there is no $2$ before the square root. $\endgroup$
    – M. Stefan
    Feb 4, 2017 at 15:47
  • $\begingroup$ it seems correct to me now $\endgroup$
    – Maczinga
    Feb 4, 2017 at 15:51

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