<s>[INST] Find $\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k$ I am trying to find the following limit: 
\begin{align}
\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}}  \sum_{k=0}^\infty  \frac{\sqrt{k+ n }}{k!} (n+a)^k
\end{align}
for some fixed $a>0$. 
Things that tired. 
We can come up with the following bound: 
\begin{align}
\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}}  \sum_{k=0}^\infty  \frac{\sqrt{k+ n }}{k!} (n+a)^k \le  \lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}}  \sum_{k=0}^\infty \left(  \frac{\sqrt{k }}{k!} (n+a)^k + \frac{\sqrt{ n }}{k!} (n+a)^k\right)\\
 \le  \lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}}     \sum_{k=0}^\infty \frac{\sqrt{k }}{k!} (n+a)^k + e^{a}
\end{align}
 [/INST] Defining $p(k, \lambda) = \frac{\lambda^k}{k!}e^{-\lambda}$, then
\begin{align*}
f(k, n) = \sqrt{1 + \frac{k}{n}}e^a \frac{(n+a)^k}{k!}e^{-(n+a)} = \sqrt{1 + \frac{k}{n}}e^ap(k, n+a)
\end{align*}
So the summation is equivalent to, for a random variable $X \sim \text{Pois}(n+a)$,
\begin{align*}
e^{a}\mathbb{E}\left[\sqrt{1 + \frac{X}{n}}\right]
\end{align*}
Since $X/n \rightarrow 1$ almost surely by the Strong Law of Large Numbers and $\mathbb{E}[\frac{X}{n}] = 1 + \frac{a}{n} \le 1 + a < \infty$, dominated convergence allows us to exchange $\mathbb{E}$ and $\lim$, and so
\begin{align*}
\lim_{n\rightarrow \infty} e^{a}\mathbb{E}\left[\sqrt{1 + \frac{X}{n}}\right] = e^{a}\mathbb{E}\left[\sqrt{1 + \lim_{n\rightarrow \infty}\frac{X}{n}}\right] = e^{a} \sqrt{2}
\end{align*}
</s>