Let $g:\mathbb N \to \mathbb N$ be a function such that $g(m) \le m$ for all $m$. Let $m$ be a large positive integer and let $0 < p < 1/2$. Define $c_m(p) := \sum_{k = g(m)}^m{m\choose k}(1-p)^{m-k}p^k$, the probability that $m$ draws of a binomial random variable with parameter $p$, result in at most $g(m)$ heads.
Question. Under what conditions on the function $g$ do we have $c_m(p) \to 0$ as $m \to \infty$ ?
Let $\cal{B}(m,p)$ denote a Binomial random variable with parameters $(m,p)$. By the CLT \begin{equation} \frac{\mathcal{B}(m,p) - mp}{\sqrt{m}} \end{equation} converges in distribution to a centered Gaussian with variance $p(1-p)$ as $m \to \infty$. By definition of $c_m(p)$ \begin{equation} \lim\limits_{m\to\infty} c_m(p) = \lim\limits_{m \to \infty} \mathbb{P}\left(\frac{\mathcal{B}(m,p) - mp}{\sqrt{m}} \geq \frac{g(m) - mp}{\sqrt{m}}\right). \end{equation} From the above display one can infer that $c_m(p) \to 0$ as $m\to\infty$ iff \begin{equation} \liminf\limits_{m \to \infty} \frac{g(m)-mp}{\sqrt{m}} = \infty. \end{equation}