Let's instead consider the sum
$$ \sum_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!} $$ which of course differs from yours just by a factor of $e^{-A}$.
Then this sum is the probability that a Poisson random variable of mean $A$ is at least $A + C\sqrt{A}$.
A Poisson with mean $A$ has standard deviation $\sqrt{A}$, and as $A \to \infty$ the Poissons become asymptotically normal. So we have
$$ \sum_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!} \to \Phi(C) $$
as $A \to \infty$, where $\Phi$ is the CDF of the standard normal. So your sum is asymptotic to $e^A \Phi(C)$.
Alternatively, if you'd like an explicit inequality, your sum can be bounded above by the geometric series with first term $A^B/B!$ and common term ratio $A/B$. Therefore, your sum is smaller then less than $A^B/B! $ {A^B \times B/A = A^{B+1}/{B-1}!$. With over B!} {1 \over 1-A/B} $$ and this can be rewritten as $$ {A^B \over B!} \left( 1 + {\sqrt{A} \over C} \right) $$ The product $A^B/B!$ is, as $A \to \infty$ with $B = A + C \sqrt{A}$ this is asymptotic to sqrt{A}$, $e^{A} e^{-c^2/2}/(c\sqrt{2\pi})$ by Stirling; $ {1 \over \sqrt{2\pi}} e^{-C^2/2} A^{-1/2} e^A (1+o(1))$$ by Stirling's formula. In the factor $1 + \sqrt{A}/C$ we can neglect $1$ as $A \to \infty$, so we get that
$$ \sum_{k = A+C\sqrt{A}}^\infty {e^{-A} A^k \over k!} \le {1 \over \sqrt{2\pi}} C e^{-C^2/2} e^A (1 + o(1)) $$
By, say, the double inequality (26) here it should be possible to get explicit bounds.
