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?

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

As $\lfloor \frac{n+1}2\rfloor <n$ for $n\ge2$, we have $\lfloor \frac{n+1}2\rfloor\le (n-1)!=\frac1n\cdot n!$

  • 1
    $\begingroup$ I think you are missing a " ! ", and how do you get to the second inequality anyway? $\endgroup$ – J.C.VegaO Sep 13 '19 at 20:11
  • $\begingroup$ It is because if $m$ and $M$ are non-negative integers with $m\le M$, then $m!\le M!$ (and $\lfloor \frac{n+1}{2}\rfloor \le n-1$). $\endgroup$ – Minus One-Twelfth Sep 13 '19 at 20:15

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