= Solution
A <Poisson distribution> with parameter $\lambda>0$ assigns <probability mass function>
$$
P(Z=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots.
$$
The <exponential series> makes these probabilities sum to one. Shifting the indices in the absolutely convergent sums gives
$$
E Z=e^{-\lambda}\sum_{k\geq1}\frac{k\lambda^k}{k!}=\lambda,
\qquad E[Z(Z-1)]=\lambda^2.
$$
Hence $E Z^2=\lambda^2+\lambda$ and
$$
\boxed{E Z=\operatorname{Var}Z=\lambda.}
$$
For <independent random variables> $Z_1,\ldots,Z_n$ of parameter one, let $S_n=\sum_jZ_j$. Summing their product <probability mass function> over all nonnegative $k_1+\cdots+k_n=k$ gives
$$
P(S_n=k)=e^{-n}\sum_{k_1+\cdots+k_n=k}\frac1{k_1!\cdots k_n!}
=e^{-n}\frac{n^k}{k!}.
$$
The final equality follows from the <multinomial theorem> applied to $(1+\cdots+1)^k$. Thus \b[$S_n$ has the Poisson distribution of parameter $n$].
Each $Z_j$ has <expectation> and <variance> one, so the <central limit theorem> gives <convergence in distribution>
$$
\frac{S_n-n}{\sqrt n}\ \Longrightarrow\ N(0,1).
$$
Zero is a continuity point of the <standard normal cumulative distribution function>. Therefore
$$
\boxed{e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=P(S_n\leq n)
=P\left(\frac{S_n-n}{\sqrt n}\leq0\right)\longrightarrow\frac12.}
$$
The discrete atom at $S_n=n$ causes no difficulty: the <central limit theorem> applies to the entire event with threshold zero.
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