我如何才能找到正选楼（N / 2）的渐进增长？我试过了 使用的扩展，并得到它等于

How can I find the asymptotic growth of n choose floor(n/2) ? I tried to use the expansion and got that it is equal to

[n*(n-1)*........*(floor(n/2)+1)] / (n-floor(n/2))!

任何想法，我怎么能去从那里？ 任何帮助是pciated AP $ P $，preFER暗示了答案

Any idea how can i go from there? Any help is appreciated, prefer hints over answers

推荐答案

使用斯特灵公式，你得到

n! = \sqrt{2n\pi}(n/e)^n

如果你代入$ \选择{N} {N / 2} $中，最终结束了

If you substitute it into $\choose{n}{n/2}$, you should eventually end up with

2^{n+1/2}/\sqrt{n\pi}

PS。你可能要检查我的数学，然后再实际使用的答案： - ）

PS. you might want to check my math before you actually use the answer :-)