# Limit of consecutive sum of binomials

Let $a$ be a positive integer. What is the limit $$\lim_{n\rightarrow\infty}\frac{\dbinom{an}{0}+\dbinom{an}{1}+\cdots+\dbinom{an}{n-1}}{2^{an}}$$ where $n$ takes on integer values?

Since the binomials can be approximated by a normal distribution when $n$ gets large, the answer should have something to do with the normal distribution probability distribution. But I'm confused what the relation is and how to derive the answer.

## 1 Answer

The probability you're interested in is $\Pr[\mathrm{Bin}(an,1/2)<n]$. There are three cases: if $a<2$ then the limit is $0$, if $a = 2$ then it is $1/2$, if $a > 2$ then it is $1$. When $a \neq 2$, you can show this using Chebyshev's theorem. The case $a = 2$ can be handled directly, since by symmetry we know that $$\frac{\binom{2n}{0} + \cdots + \binom{2n}{n-1} + \frac{1}{2} \binom{2n}{n}}{2^{2n}} = \frac{1}{2},$$ while $\binom{2n}{n}/2^n = O(1/\sqrt{n})$.