Calculating $\lim\limits_{C \rightarrow \infty} -\frac{1}{C} \log(1 - p \sum\limits_{k=0}^{C}e^{-\gamma C} (\gamma C)^k/k!) $ How to calculate the following limit:
$$\lim_{C\rightarrow \infty} -\frac{1}{C} \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right)$$
Given that $0 \leq \gamma \leq 1$ and $0 \leq p \leq 1$. At least any tips about approaching the solution!
 A: Note that $\mathrm e^{-\gamma C}\sum\limits_{k=0}^{C}\frac1{k!}(\gamma C)^k=\mathrm P(X_{\gamma C}\leqslant C)$, where $X_{\gamma C}$ denotes a Poisson random variable with parameter $\gamma C$. In particular, if $p\lt1$, the argument of the logarithm is between $1-p$ and $1$. Likewise, if $\gamma=1$, a central limit argument yields $\mathrm P(X_{C}\leqslant C)\to\frac12$. In both cases, the limit is zero.
From now on, assume that $p=1$ and that $\gamma\lt1$. One is interested in 
$$
1-\mathrm e^{-\gamma C}\sum\limits_{k=0}^{C}\frac1{k!}(\gamma C)^k=\mathrm P(X_{\gamma C}\gt C).
$$
Introduce some i.i.d. Poisson random variables $\xi$ and $\xi_k$ with parameter $1$ and, for every positive integer $n$, $\eta_n=\xi_1+\cdots+\xi_n$. Then, on the one hand $\eta_n$ is a Poisson random variable of parameter $n$ and on the other hand, for every $t\gt1$, the behaviour of $\mathrm P(\eta_n\gt tn)$ is described by a large deviations principle. More precisely,
$$
\mathrm P(\eta_n\gt tn)=\mathrm e^{-nI(t)+o(n)},\quad\text{where}\ 
I(t)=\max\limits_{x\geqslant0}\left(xt-\log\mathrm E(\mathrm e^{x\xi})\right).
$$
In the present case, $\log\mathrm E(\mathrm e^{x\xi})=\mathrm e^x-1$ hence $I(t)=t\log t-t+1$ for every $t\gt1$. Using this result for $n=\lfloor\gamma C\rfloor$ and $t=1/\gamma$, one gets
$$
\lim\limits_{C\to+\infty}-\frac1C\log\mathrm P(X_{\gamma C}\gt C)=\gamma I(1/\gamma)=\gamma-1-\log\gamma.
$$
A: Perhaps not what you wanted to ask ...
$$\begin{align}
&0 \le \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!} \le 1,
\\
&\log(1-p)
\le \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right) \le \log 1 = 0
\\
&\lim_{C \rightarrow \infty} -\frac{1}{C} \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right) = 0
\end{align}$$
