Checking if Tedious Limit is $0$.

Question

I am trying to rigorously (without a calculator) show $$\lim_{n\to \infty }\sqrt{n}\color{blue}{{{n}\choose {\Big[ np + \sqrt{np(1-p)}\,\Big]}}p^{\Big[np + \sqrt{np(1-p)} \,\Big]}{(1-p)^{\bigg(n-\Big[np + \sqrt{np(1-p)}\,\Big]\bigg)}}}=0$$ where $0

https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem

I am not completely convinced that the limit is 0, but after seeing some computations on Mathematica

https://mathematica.stackexchange.com/questions/134523/why-wont-limit-evaluate-and-what-can-be-done-about-it?noredirect=1#comment362965_134523 ,

I see that it tends to $0$ for several rational values of $p$. I want to see whether the limit is $0$ for all $p$ in $(0,1)$.

What I've tried: (1) Using ${{a}\choose{b}}\le (\frac{ae}{b})^b$ and dropping $[ \cdot]$ for simplicity (and hopefully at no cost) to get $$\sqrt{n}\Bigg(\frac{ne}{np + \sqrt{np(1-p)}}\Bigg)^{np + \sqrt{np(1-p)}}p^{np + \sqrt{np(1-p)}}(1-p)^{n-np + \sqrt{np(1-p)}} \\\le \sqrt{n}\Bigg(\frac{nep}{np(1-p)+n\sqrt{\frac{p(1-p)}{n}}(1-p) }\Bigg)^{np+ \sqrt{np(1-p)}}(1-p)^n\\\le \sqrt{n}\bigg(\frac{e}{(1-p)}\bigg)^{2np}(1-p)^n$$

but according to Mathematica the latest expression doesn't tend to 0, so I may need a better upper bound.

Answer 1

Excavating. This does not hold. Actually the limit is $\dfrac{1}{\sqrt{2\pi ep(1-p)}}$.

Let $p_n=[np+\sqrt{np(1-p)}]/n$ ($\to p$ when $n\to\infty$); the quantity in question is $$ x_n=\sqrt{n}\binom{n}{np_n}\color{blue}{p}^{np_n}(1-\color{blue}{p})^{n(1-p_n)}. $$ Now let $$ y_n=\sqrt{n}\binom{n}{np_n}\color{red}{p_n}^{np_n}(1-\color{red}{p_n})^{n(1-p_n)}; $$ the Stirling's formula says $\displaystyle\lim_{n\to\infty}y_n=\frac{1}{\sqrt{2\pi p(1-p)}}$. Further, $x_n/y_n=e^{nf(p,p_n)}$, where $$\begin{gathered}f(p,t)=t\ln\frac{p}{t}+(1-t)\ln\frac{1-p}{1-t}, \\ \frac{\partial\!f}{\partial t}=\ln\frac{p}{1-p}-\ln\frac{t}{1-t}, \quad \frac{\partial^2\!f}{\partial t^2}=-\frac{1}{t(1-t)}, \\ p_n=p+\sqrt{\frac{p(1-p)}{n}}+O\Big(\frac{1}{n}\Big),\end{gathered}$$ which gives $nf(p,p_n)=-1/2\ +\ O(n^{-1/2})$ and the claim above.