Asymptotics of a sum involving factorials

Question

I'm interested in the large-$n$ behavior of the sum $$a(m,n) = \sum_{k=0}^m \frac{x^k}{k! (n-k)!}$$

Answer 1

$$a_n= \frac{(x+1)^n}{n!}$$ Use the Stirling formula, for $n\to+\infty$ $$a_n \sim \frac{1}{\sqrt{2\pi n}} \left(\frac{e(x+1)}{n} \right)^n $$

Answer 2

For the new question, here is the answer.

Suppose $x>0$. Let's define $p= \frac{x}{x+1}$, then $x = \frac{p}{1-p}$. We have

$$\begin{align} a(m,n) &=\sum_{k=0}^m \frac{x^k}{k! (n-k)!}\\ &=\sum_{k=0}^m \frac{p^m}{(1-p)^m}\cdot \frac{1}{k!(n-k)!}\\ &=\frac{1}{n!(1-p)^n}\color{red}{\sum_{k=0}^m \frac{n!}{k!(n-k)!}\cdot p^m(1-p)^{n-m}}\\ \end{align}$$

Let's define $Y_n$ following the binominal distribution $B(n,p)$, then the term in red color is exacly the probability $P(Y_n\le m)$.

We know that the binominal distribution $B(n,p)$ is a sum of $n$ i.i.d Bernoulli random variable, then

$$P(Y_n\le m) = P(\sum_{k=1}^nX_k \le m) = P(\frac{1}{n}\sum_{k=1}^nX_k \le \frac{m}{n})$$

According to the central limit theorem, for $n$ large, the random variable $\frac{1}{n}\sum_{k=1}^nX_k$ follow the normal distribution $\mathcal{N}(p,p(1-p))$, then $$\begin{align} P(Y_n\le m) \sim P(\mathcal{N}(p,p(1-p))\le \frac{m}{n})&= P\left(\mathcal{N}(0,1)\le \frac{1}{\sqrt{p(1-p)}} \left(\frac{m}{n} -p \right)\right)\\ &=\Phi\left(\frac{1}{\sqrt{p(1-p)}} \left(\frac{m}{n} -p \right)\right) \end{align}$$ with $\Phi(.)$ the cumulative distribution function of the standard normal distribution.

Finally, for $n$ large, using the Stirling's formula, we have the following approximation

$$\begin{align} a(m,n) &\sim \frac{1}{n!(1-p)^n}\Phi\left(\frac{1}{\sqrt{p(1-p)}} \left(\frac{m}{n} -p \right)\right)\\ &\sim \color{blue}{ \frac{1}{\sqrt{2\pi n}}\left(\frac{e(x+1)}{n}\right)^n\Phi\left(\frac{x+1}{\sqrt{x}} \left(\frac{m}{n} -\frac{x}{x+1} \right)\right)} \end{align}$$

Q.E.D