I have a problem that I'm unsure how to solve with a method suggested by the question hint.
Problem A particle takes a random walk on the integers, starting at 0. At each step it advances to the next integer (move to the right) with probability 1/4 or goes back to the previous integer (moves to the left) with probability 3/4. Let the position of the particle after $n$ moves be $X_n$. Derive an asymptotic equivalent for $P(X_{2n} = 0)$. (Hint: use Stirling's approximation.)
As you can tell it is a very basic random walk problem, but I'm not familiar with Stirling's method. It seems to be using $In(x)$ integral to derive a curvature approx. for the probability. I think I have to use this equation at some point: $$In(x)!=nIn(n)-n+1, Interval(1,n)$$ Would like to have some guidance on applying it to the problem.
$$\mathbb{P}(X_{2n} =0) = \mathrm{Binomial}\left(n;2n,\frac{1}{4}\right) = \binom{2n}{n}\left(\frac{3}{16}\right)^n$$ Using Stirling's formula $$ \binom{2n}{n} \approx \frac{4^n}{\sqrt{n\pi}}$$ So $$ \mathbb{P}(X_{2n} =0) \approx \frac{1}{\sqrt{n\pi}}\left(\frac{3}{4}\right)^n$$