Calculating $\lim\limits_{C \rightarrow \infty} -\frac{1}{C} \log(1 + (e^{-p \gamma C} - 1) \sum\limits_{k=0}^{C}e^{-\gamma C} (\gamma C)^k/k!) $ How to calculate the following limit:
$$\lim_{C\rightarrow \infty} -\frac{1}{C} \log\left(1 + (e^{-p \gamma C} - 1)  \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right)$$
Given that $0 \leq \gamma \leq 1$ and $0 \leq p \leq 1$. This is a modified version of a problem I've posted earlier here.
 A: The argument in the logarithm is $A(C)=\mathrm e^{-p\gamma C}+(1-\mathrm e^{-p\gamma C})\mathrm P(X_{\gamma C}\geqslant C)$ where $X_{\gamma C}$ is a Poisson random variable with parameter $\gamma C$. 
Assume that $\gamma\lt1$. When $C\to\infty$, $[X_{\gamma C}\geqslant C]$ is a large deviations event and 
$$
\mathrm P(X_{\gamma C}\geqslant C)=\mathrm e^{-CI(\gamma)+o(C)}
$$ 
for some positive $I(\gamma)$ I will let you discover. Thus, 
$\log A(C)=-\inf\{p\gamma,I(\gamma)\}C+o(C)$ 
and the limit you are after is: $\inf\{p\gamma,I(\gamma)\}$. 
When $\gamma=1$, $I(\gamma)=0$ and the limit is $0$ for every $p$.
Edit: The large deviations estimate mentioned above is deduced from the usual exponential inequality. Namely, for every nonnegative $t$,
$$
\mathrm P(X_{\gamma C}\geqslant C)\leqslant\mathrm e^{-tC}\cdot\mathrm E(\mathrm e^{tX_{\gamma C}})=\mathrm e^{-tC}\cdot\mathrm e^{\gamma C(\mathrm e^t-1)}.
$$
Hence, for every $\gamma\lt1$, the estimate above holds with
$$
I(\gamma)=\sup\limits_{t\gt0}\left(t-\gamma(\mathrm e^t-1)\right)=\gamma-1-\log\gamma.
$$
A: Another thought: (I use $(n,a,b)$ instead of $(C,\gamma,p)$)
Claim 1 $\hspace{5ex}$ Let $a_n,b_n>0$ with $a_n,b_n=\mathcal O(n^p)$ for some $p\in\mathbb R$ and $a,b>0$. Then $$\lim_{n\to+\infty}-\frac{1}{n}\ln\left(1+(a_ne^{-an}-1)(1-b_ne^{-bn})\right)=\min\{a,b\}.$$
Indeed $\hspace{5ex}$ For $a\leq b$, (the other case is similar) $$\begin{aligned}-\frac{1}{n}\ln\left(1+(a_ne^{-an}-1)(1-b_ne^{-bn})\right)&=a-\frac{\ln a_n}{n}-\frac{1}{n}\ln\left(1+b_ne^{-(b-a)n}-a_nb_ne^{-bn}\right)\\ \stackrel{a_n,b_n=\mathcal O(n^p)}{=}&a+o(1)\hspace{40ex}\Box\end{aligned}$$
$\bullet$ If $a=0$ or $b=0$ the quantity equals $0$.
$\bullet$ For $a,b\neq0$ and $a\neq1$, from Taylor's theorem we can write
$$\sum_{k=0}^{n}\frac{(an)^k}{k!}=e^{an}-\frac{1}{n!}\int_{0}^{1}e^t(an-t)^n\,dt:=e^{an}-A_n,$$ 
and
$$A_n\stackrel{t=any}{=}\frac{(an)^{n+1}}{n!}\int_{0}^{1}e^{any}(1-y)^n\,dy=\frac{(an)^{n+1}}{n!}\mathcal O(1),$$
so
$\begin{aligned}e^{-an}\sum_{k=0}^{n}\frac{(an)^k}{k!}&=e^{-an}\left(e^{an}-\frac{(an)^{n+1}}{n!}\mathcal O(1)\right)\\&\stackrel{(1)}{=}1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\mathcal O(1)\\&=1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\end{aligned}$
Now the given quantity equals
$$-\frac{1}{n}\ln\left(1+\left(e^{-abn}-1\right)\left(1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\right)\right)$$
and since $1-a+\ln a<0$ from Claim 1 the quantity goes to $\min\{a-1-\ln a,ab\}$.

$(1)$ From Stirling's formula

$\bullet$ For $a=1$ and $b\neq0$, proceeding as above, we need a better estimate for $\int_{0}^{1}e^{ny}(1-y)^n\,dy$ than just $\mathcal O(1)$.
Claim 2 $\hspace{5ex}$ For $a>-1$ and $c>0$ it is $$\int_{0}^{c}t^ae^{-nt}\,dt = \frac{\Gamma(a+1)}{n^{a+1}}+\mathcal O(n^{-1}e^{-nc}).$$
Indeed $\hspace{3ex}$ for $a>-1$ and $c>0$ :
$\begin{align}\int_{0}^{c}t^ae^{-nt}\,dt &= \int_{0}^{+\infty}t^ae^{-nt}\,dt-\int_{c}^{+\infty}t^ae^{-nt}\,dt\notag\\
 &\stackrel{nt=x}{=} \frac{\Gamma(a+1)}{n^{a+1}}-\frac{1}{n^{a+1}}\int_{nc}^{+\infty}x^ae^{-x}\,dx\notag\\
 &\stackrel{x=(1+u)nc}{=} \frac{\Gamma(a+1)}{n^{a+1}}-c^{a+1}\int_{0}^{+\infty}(1+u)^ae^{-(1+u)nc}\,du\notag\\
 &\stackrel{1+u\leq e^u}{\Rightarrow} \notag\\
\left|\int_{0}^{c}t^ae^{-nt}\,dt-\frac{\Gamma(a+1)}{n^{a+1}}\right| &\leq c^{a+1}e^{-nc}\int_{0}^{+\infty}e^{u(a-nc)}\,dt=\mathcal O(n^{-1}e^{-nc})\notag\\
&\Longrightarrow  \notag\\
\int_{0}^{c}t^ae^{-nt}\,dt &= \frac{\Gamma(a+1)}{n^{a+1}}+\mathcal O(n^{-1}e^{-nc}).\hspace{25ex}\Box\notag
\end{align}$
Now looking at
$$\int_{0}^{1}e^{nx}(1-x)^n\,dx$$
we set $\phi(x)=-x-\ln(1-x)$ so
$$\frac{1}{\phi^{'}(x)}=-1+\frac{1}{x}$$
but
$$t:=\phi(x)=\frac{x^2}{2}+\mathcal O(x^3)=\frac{x^2}{2}\left(1+\frac{2x}{3}+\mathcal O(x^2)\right)$$
so, as $t\geq0$, setting $u=\sqrt{2t}$ we get
$$u=x\left(1+\frac{2x}{3}+\mathcal O(x^2)\right)^{1/2}=x+\frac{x^2}{3}+\mathcal O(x^3)$$
and reverse setting $x=u+a_2u^2+\mathcal O(u^3)$ and equating coefficients to get
$$x=u-\frac{1}{3}u^2+\mathcal O(u^3).$$
This will give
$$\frac{1}{\phi^{'}(x)}=\frac{1}{\phi^{'}\left(\phi^{-1}(t)\right)}=\frac{\sqrt{2}}{2}t^{-1/2}-\frac{2}{3}+\mathcal O(t^{1/2})$$
and applying Claim 2
$$\int_{0}^{1}e^{ny}(1-y)^n\,dy=\frac{\Gamma(1/2)}{\sqrt{2}}n^{-1/2}+\mathcal O(n^{-1})=\sqrt{\frac{\pi}{2}}n^{-1/2}+\mathcal O(n^{-1}).$$
With Stirling's formula again
$\begin{aligned}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}&=1-\left(\frac{1}{\sqrt{2\pi}}n^{1/2}+\mathcal O(n^{-1/2})\right)\left(\sqrt{\frac{\pi}{2}}n^{-1/2}+\mathcal O(n^{-1})\right)\\&=\frac{1}{2}+\mathcal O(n^{-1/2})\end{aligned}$
so the given quantity equals $-\frac{1}{n}\cdot\mathcal O(1)\to0$.
