My ultimate goes is to show that $P(\tau = n) = \frac{1}{2^{n-1}} $ ${n}\choose{\frac{n+x}{2}}$ $\to 0$ as $n \to \infty$. I'm trying to use Stirling's approximation: ${n}\choose{k}$ $\approx \frac{n^k}{k!}$.
This is what I have so far, ${n}\choose{\frac{n+x}{2}}$ $\approx \frac{n^{(\frac{n+x}{2})}}{(n+\frac{x}{2})!} \Rightarrow \lim_{n \to \infty} P(\tau = n) \approx \frac{1}{2^{n-1}}\frac{n^{(\frac{n+x}{2})}}{(n+\frac{x}{2})!}$ I know that $\frac{1}{2^{n-1}}$ approaches zero as $n \to \infty$. Is there a way to show that the rate of growth of the denominator has to be faster than the numerator so that this function approaches zero in the limit? I'm not quite sure how to proceed.
Best Answer
As @Tuvasbien commented, rewrite $$P = \frac{1}{2^{n-1}}{{n}\choose{\frac{n+x}{2}}}=\frac{1}{2^{n-1}}\frac{n!}{\left(\frac{n+x}{2}\right)!\left(\frac{n-x}{2}\right)!}$$ Take logarithms and use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \log (2 \pi p)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ Apply it three times and continue with Taylor series to obtain $$\log(P)=\frac{1}{2} \log \left(\frac{8}{\pi n}\right)-\frac{2 x^2+1}{4 n}+O\left(\frac{1}{n^2}\right)$$ $$P=e^{\log(P)}=2 \sqrt{\frac{2}{n\pi }}-\frac{2 x^2+1}{n\sqrt{2\pi n} }+O\left(\frac{1}{n^2}\right)$$