# 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}

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*}