= Solution
For $\lambda\geq0$, a <Poisson distribution> has mass $\mathbb P(X=k)=e^{-\lambda}\lambda^k/k!$ for $k=0,1,\ldots$; at $\lambda=0$ it is concentrated at zero. Its <moment-generating function> is
$$
\boxed{M_X(s)=\sum_{k\ge0}e^{sk}e^{-\lambda}\frac{\lambda^k}{k!}=\exp[\lambda(e^s-1)],\qquad s\in\mathbb R.}
$$
For independent $X,Y$, the <moment-generating function> of their sum is the product, $\exp[(\lambda+\mu)(e^s-1)]$. Uniqueness of the <moment-generating function> near zero therefore gives \b[$X+Y\sim\operatorname{Poisson}(\lambda+\mu)$]. Repeating the argument gives \b[$S_n=\sum_{i=1}^nX_i\sim\operatorname{Poisson}(n)$] when every parameter is one.
The displayed sequence is $a_n=\mathbb P(S_n\leq n)$. The classical <central limit theorem> states that independent identically distributed variables with finite mean $m$ and nonzero finite <variance> $\sigma^2$ have standardized sum $(S_n-nm)/(\sigma\sqrt n)$ converging in distribution to the standard <normal distribution>. Here $m=\sigma^2=1$, so
$$
\boxed{a_n=\mathbb P\left(\frac{S_n-n}{\sqrt n}\leq0\right)\longrightarrow\Phi(0)=\frac12.}
$$
Convergence of the distribution functions at the continuity point zero justifies the last step. This is the <Poisson cumulative probability at its growing mean>.
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