Suppose that
- $(k_m)_{m\in \mathbb{N}}$ is a sequence of natural numbers such that $k_m \to\infty$ and $\frac{k_m}{m} \to0$ as $m \to \infty$;
- $(p_m)_{m \in \mathbb{N}} \subset(0,1)$ is such that $p_m \to 0$ and $m p_m \to \infty$ as $m \to \infty$.
Is it true that \begin{equation} \sum_{j=0}^{k_m}\binom{m}{j}p_m^j(1-p_m)^{m-j} \to 0,~ m \to \infty? \end{equation}
I tried something along the following path (all the estimates are far from being tight, but it is just to get a sense of what I've done). From the answer to this question I know that if for example $k_m + \frac{1}{2} \le \frac{1}{3}\frac{m p_m}{\log(m p_m)}$, then there exists a constant $C>0$ such that \begin{align*} \sum_{j=0}^{k_m}\binom{m}{j}p_m^j(1-p_m)^{m-j} &\le C \binom{m}{k_m}p_m^{k_m}\sqrt{mp_m}(1-p_m)^{m-k_m} \\ &\le C(mp_m)^{k_m+\frac{1}{2}}(\exp(-(m-k_m)p_m)) \\ &=C\exp\Big(-mp_m +k_mp_m+(k_m +\frac{1}{2})\log(mp_m)\Big) \\ &\le C\exp\Big(-\frac{1}{6}mp_m \Big) \to 0,~ m\to\infty. \end{align*}
However, I'm unsatisfied since what I want is proving the result without any further assumption about how $m p_m$ and $k_m$ grow with respect to each other (or showing a counter-example if it is false, and in this case find some tight relationship between $k_m$ and $p_m$ in order to get that the result still holds). Any ideas?
Setting $p=p_m, k=k_m$, let $Z$ be a Binomial random variable with mean $mp$ and variance $mp(1-p)$. Then: \begin{align} \sum_{j=0}^{k}\binom{m}{j}p^j(1-p)^{m-j} &= prob(Z \leq k) = prob(Z/m \leq k/m) = 1 - prob(Z/m > k/m)\\ &\geq \frac{k-pm}{k}, \end{align} using Markov inequality. Assume now $k=2pm$ holds for every $m$. Then: \begin{align} \sum_{j=0}^{k}\binom{m}{j}p^j(1-p)^{m-j} &\geq \frac{1}{2}. \end{align} This shows that limit, provided it exists, is not $0$ in general without requiring further conditions on $k,p$.