 Solution

A Poisson distribution with parameter $\lambda >0$ assigns probability mass functionThe exponential series makes these probabilities sum to one. Shifting the indices in the absolutely convergent sums givesHence $E{Z}^2={\lambda }^2+\lambda$ andFor independent random variables $Z_1,\dots ,Z_n$ of parameter one, let $S_n={\sum }_j{Z}_j$. Summing their product probability mass function over all nonnegative $k_1+\cdots +k_n=k$ givesThe final equality follows from the multinomial theorem applied to $(1+\cdots +1{)}^k$. Thus $S_n$ has the Poisson distribution of parameter $n$.

Each $Z_j$ has expectation and variance one, so the central limit theorem gives convergence in distributionZero is a continuity point of the standard normal cumulative distribution function. ThereforeThe discrete atom at $S_n=n$ causes no difficulty: the central limit theorem applies to the entire event with threshold zero.