# Find $\lim_{n\to\infty}\frac{\left\lfloor \frac{n+1}{2} \right\rfloor!}{n!}$

Find $$\lim_{n\to\infty}\frac{\left\lfloor \dfrac{n+1}{2} \right\rfloor!}{n!}$$

I tried Stirling's formula but I seem to get nowhere. How should I proceed?

• $[(n+1)/2]< (n+1)/2+1<(n-1)$ for $n>5$ – kingW3 Sep 13 '19 at 20:53

As $$\lfloor \frac{n+1}2\rfloor for $$n\ge2$$, we have $$\lfloor \frac{n+1}2\rfloor\le (n-1)!=\frac1n\cdot n!$$
• It is because if $m$ and $M$ are non-negative integers with $m\le M$, then $m!\le M!$ (and $\lfloor \frac{n+1}{2}\rfloor \le n-1$). – Minus One-Twelfth Sep 13 '19 at 20:15