# Showing that $\lim\limits _{n\to\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}=0$ but the matching series does not converge

I want to show that: $$\lim\limits _{n\to\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}=0$$ And also $${\displaystyle \sum_{n=1}^{\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}}$$ Does not converge. I'm not particularly good with this sort of approximation so I would really appreciate some help.

• First, make life easy by assuming $n$ is even. Second, I bet they want you to use Stirling's formula en.wikipedia.org/wiki/Stirling%27s_approximation – Stephen Montgomery-Smith Nov 18 '13 at 4:42
• I tried opening up the binomial coefficient and doing some approximations from above of the result but then I figure that ${n \choose \left\lceil \frac{n}{2}\right\rceil }$ goes to infinity rather than $1$ like I had hoped. So you definitely need to incorporate that $2^{-n}$ in the process. I haven't tried using Stirling though. – Serpahimz Nov 18 '13 at 4:48

These are the central binomial coefficients. You can see the wiki page for the estimates $$\frac{2^n}{n+1}\le {n\choose \lceil \frac{n}{2}\rceil} \le \frac{2^n}{\sqrt{1.5n+1}}$$
Multiply through by $2^{-n}$ and you get your terms. The upper bound gives the first part; the lower bound gives the second part by comparison with the harmonic series.
• How would you formally prove those bounds though? I would be happy to see how they are obtained from the original form of ${n \choose \left\lceil \frac{n}{2}\right\rceil }$ – Serpahimz Nov 18 '13 at 4:52