I'm reading a derivation of the diffusion equation in JD Murray's Mathematical Biology based on a simple random walk model:
A particle moves randomly backward and forward along a line in fixed steps $\Delta x$ that are taken in a fixed time $\Delta t$. If the motion is unbiased then it is equally probable that the particle takes a step to the right or left. After time $N \Delta t$ the particle can be anywhere from $-N \Delta x$ to $N \Delta x$ if we take the starting point of the particle as the origin.
Let $p(m,n) := $ the probability that a particle reaches a point $m$ space steps to the right (that is, to $x = m \Delta x$) after $n$ time steps (that is, after a time $n \Delta t$). It is shown that \begin{align*} p(m,n) = \frac{1}{2^n} \frac{n!}{a! (n-a)!}, \quad \text{where } a := \frac{n+m}{2} \end{align*} for $|m| \leq n$. The author then says that using Stirling's formula $$ n! \sim (2\pi n)^{1/2} n^n e^{-n} $$
and "a little algebra" we get
$$ p(m,n) \sim \left(\frac{2}{\pi n}\right)^{1/2} e^{-m^2/(2n)}, \qquad m \gg 1, \quad n \gg 1. $$
I am just trying to verify this claim. Here is my attempt:
\begin{align*} p(m,n) &\sim \frac{1}{2^n} \frac{\sqrt{2\pi n} \, n^n e^{-n}} {\sqrt{2\pi a} \, a^a e^{-a} \sqrt{2 \pi (n-a)} \, (n-a)^{n-a} e^{-(n-a)}} \\[5pt] &= \frac{1}{2^n} \frac{\sqrt{n}}{\sqrt{a} \sqrt{2 \pi (n-a)}} n^n e^{-n} a^{-a} e^a (n-a)^{-(n-a)} e^{n-a} \\[5pt] &= \frac{1}{2^n} \frac{\sqrt{n}}{\sqrt{2 a \pi (n-a)}} n^n a^{-a} (n-a)^{-n} (n-a)^{a} \\[5pt] &= \frac{1}{2^n} \frac{\sqrt{n}}{\sqrt{2 a \pi (n-a)}} \left(\frac{n}{n-a}\right)^n \left(\frac{n-a}{a}\right)^a. \end{align*}
Now we have \begin{align*} \left( \frac{n}{n-a} \right)^a = \left( \frac{n-a}{n} \right)^{-a} = \left[\left(1 - \frac{a}{n} \right)^{a}\right]^{-1} \to (e^{-a})^{-1} = e^a \quad \text{ as } n \to \infty. \end{align*}
Hence, \begin{align*} p(m,n) \sim \frac{1}{2^n} \frac{\sqrt{n}}{\sqrt{2 a \pi (n-a)}} e^a \left(\frac{n-a}{a}\right)^a \end{align*}
And here I am not sure how to proceed...Any help would be appreciated.
Replacing $a=\frac{n+m}{2}$ and using the gamma function instead of the factorial $$p(m,n) =\frac 1{2^n} \frac{\Gamma (n+1) } { \Gamma \left(\frac{n-m+2}{2} \right)\,\, \Gamma \left(\frac{n+m+2}{2} \right)}$$
Taking logarithms and using Stirling approximation $$\log[p(m,n)]=-\frac{1}{2} (\log (n)+\log (2 \pi )-2 \log (2))-\frac{2 m^2+1}{4n}+\frac{m^2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$p(m,n)=e^{\log[p(m,n)] }=\sqrt{\frac{2}{\pi n}}\exp\left(-\frac{2 m^2+1}{4n} +O\left(\frac{1}{n^2}\right) \right)$$
Since $m\gg 1$ then $$\color{blue}{\large p(m,n) \sim \sqrt{\frac{2}{\pi n}}\,\,e^{-\frac{m^2}{2 n}}}$$