Necessary and sufficient conditions on $C>0$ and $p \in (0, 1)$ such that $\sum_{k=\lceil pm\rceil}^m{m\choose k}e^{-Ck} \to 0$ as $m \to \infty$ Let $m$ be a large positive integer, $C>0$ and $0 < p < 1$. Define $s_m(p,C) := \sum_{k=\lceil pm\rceil}^m{m\choose k}e^{-Ck}$.
Question. What are sufficient and necessary conditions on $p$ and $C$ such that $s_m(p,C) \to 0$ as $m\to\infty$ ?
Observations
Define $H(p) := -p\log(p) - (1-p)\log(1-p)$. Then
$$
s_m(p,C) \le \sum_{k \ge pm}{m\choose k} \max_{k \ge pm}e^{-Ck} \le e^{H(p)m}e^{-Cm} = e^{-pG(p)m},
$$
where $G_C(p) := C-H(p)/p$. Now, one computes $G'_C(p) = -\log(1-p)/p^2 > 0$, and so $G_C$ is strictly increasing on $(0, 1)$. In particular, $G_C(p) > 0$ iff $p > G_C^{-1}(0)$. Thus, $s_m(p,C) \le e^{G(p)m} \to 0$ if $p > G_C^{-1}(0)$. Though sufficient, there is no reason this condition "$p>G_C^{-1}(0)$" is necessary to ensure $s_m(p,C) \to 0$.
 A: I had tried to approach this by approximating the sum as a Binomial Distribution, but as the range of this summation leaves the range of a fixed number of standard deviations, that approximation was not really applicable.

Bounding by a Geometric Sequence
Note that for $\lceil pm\rceil\le k\lt m$,
$$
\frac{\binom{m}{k+1}}{\binom{m}{k}}=\frac{m-k}{k+1}\le\frac{m-\lceil pm\rceil}{\lceil pm\rceil}\le\frac{1-p}p\tag1
$$
Induction and $(1)$ show that for $\lceil pm\rceil\le k\lt m$,
$$
\binom{m}{k}e^{-Ck}
\le\binom{m}{\lceil pm\rceil}e^{-C\lceil pm\rceil}\left(\frac{1-p}pe^{-C}\right)^{k-\lceil pm\rceil}\tag2
$$
where $(2)$ is an equality when $k=\lceil pm\rceil$.
Stirling gives
$$
\binom{m}{\lceil pm\rceil}e^{-C\lceil pm\rceil}
\sim\frac1{\sqrt{2\pi p(1-p)m}}\left(\frac{e^{-Cp}}{p^p(1-p)^{1-p}}\right)^m\tag3
$$
If
$$
C\ge-\frac1p\log\left(p^p(1-p)^{1-p}\right)\tag4
$$
then $\frac{e^{-Cp}}{p^p(1-p)^{1-p}}\le1$, and $(3)$ will tend to $0$ as $m\to\infty$. Inequality $(4)$ also says that $\frac{1-p}pe^{-C}\le(1-p)^{1/p}\lt1$ and therefore, the sum in $k$ of $(2)$ converges.

Conclusion
If $(4)$ holds, then
$$
\sum_{k\ge\lceil pm\rceil}\binom{m}{k}e^{-Ck}
\lesssim\frac1{\sqrt{2\pi p(1-p)m}}\left(\frac{e^{-Cp}}{p^p(1-p)^{1-p}}\right)^m\frac1{1-\frac{1-p}pe^{-C}}\tag5
$$
and the right hand side of $(5)$ tends to $0$ as $m\to\infty$.
If $(4)$ does not hold, then even $(3)$, the first term in the sum, will grow without bound as $m\to\infty$.

Graph of the Lower Bound for $\boldsymbol{C}$

