Limit of Binomial CDF Let $\operatorname{Binom}(n, p)$ be $n$ trials with probability of success $p$. I want to find 
$$\lim\limits_{n \to \infty} P(\operatorname{Binom}(n, p) \geq n/2)$$
I didn't know how to do this with the binomial distribution itself, so I tried transforming it into a random walk. That is, if we have $A_i = 1$ with probability $p$ and $A_i = -1$ with probability $1-p$, this is equivalent to 
$$\lim\limits_{n \to \infty} P\left(\sum\limits_{i = 1}^n A_i \geq 0\right)$$ 
but I wasn't sure where to go from here either.
 A: Note that $\operatorname{Var}(\operatorname{Binom}(n,p))=np(1-p)$, and so you can use Chebyshev's inequality to show that 
$$P\left(\left|\frac{\operatorname{Binom}(n,p)}{n}-p\right|>\epsilon\right)\to 0$$
for any $\epsilon>0$.
This immediately gives the answer $0$ for any $p<1/2$ and $1$ for any $p>1/2$. For $p=1/2$ you also need to check that $P(\operatorname{Binom}(n,1/2)=n/2)\to 0$; one way to do this is by Stirling's approximation.
A: Using your random walk idea for $p\neq 0.5$, you can see that $\sum_{i=1}^n A_i \geq 0$ iff $\dfrac{\sum_{i=1}^n A_i}{n}\geq 0$. Since the ${A_i}$ are iid distributed according to the law $2X-1$ where $X$ is Bernoulli distributed with parameter $p$, you have the almost sure convergence $\dfrac{\sum_{i=1}^n A_i}{n}\to 2\mathbb{E}[X]-1 = 2p-1$. This implies the convergence in probability to the same limit, such that $\lim_{n\to\infty}\mathbb{P}[\sum_{i=1}^n A_i \geq 0 ] = 0$ if $p<0.5$ since $2p-1<0$ and vice versa.
The case $p=0.5$ can be done via Sterling's formula.
A: Binomial is a generalization of Bernoulli to $n$ experiments. Consider rv 
$$
V_n = S_n - \frac{n}{2}
$$
where $S_n = \sum_{k=1}^{n} X_n, X_k \sim Bernoulli (p)$, so $\mathbf{E}V_n = n(p-\frac{1}{2}), \mathbf{Var}V_n = \mathbf{Var}S_n = np(1-p)$. For large $n$ you can use CLT:
$$
P(V_n \geq 0) = 1-P(V_n <0) = P\bigg(\frac{V_n - n(p-\frac{1}{2})}{n\sqrt{p(1-p)}} \leq \frac{\frac{1}{2}-p}{\sqrt{p(1-p)}}\bigg) = \Phi\bigg(\frac{\frac{1}{2}-p}{\sqrt{p(1-p)}}\bigg)
$$ 
where $\Phi$ is CDF of standard normal distribution
