Finding $\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}.$ Can someone please help solve part (iv) of the problem? I am lost on how to do it. The following work before it may serve to help. Thanks.
A random variable $X$ has a Poisson distribution with parameter $\lambda > 0$ if $X$ takes on only nonnegative integer values and $$\mathbb{P}\{X=k\} = e^{-\lambda}\frac{\lambda^k}{k!}, \hspace{20 pt} k =0,1,2,\dots.$$ In this problem, suppose that $X,Y$ are independent Poisson random variables on a probability space with parameters $\lambda_X, \lambda_Y,$ respectively.
(i) Find $\mathbb{E}[X], \text{Var}[X],\mathbb{E}[Y], \text{Var}[Y].$
(ii) Show that $X+Y$ has a Poisson distribution with parameter $\lambda_X + \lambda_Y$ by computing directly $$\mathbb{P}\{X+Y = k\}.$$
(iii) Find the characteristic function for $X,Y$ and use this to find an alternative derivation of (ii).
(iv) Suppose that for each integer $n>\lambda$ we have independent random variables $Z_{1,n}, Z_{2,n},\dots, Z_{n,n}$ with $$\mathbb{P}\{Z_{j,n} = 1\} = 1 - \mathbb{P}\{Z_{j,n} = 0\} = \frac{\lambda}{n}.$$
Find $$\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}.$$
$\textit{Proof.}$ Give that $\displaystyle{\mathbb{P}\{X=k\} = e^{-\lambda}\frac{\lambda^k}{k!}, k =0,1,2,\dots},$ $X\sim \text{ Poisson}(\lambda X),$ and $Y\sim \text{ Poisson}(\lambda Y),$ then $$\mathbb{E}[X] = \sum_{k=0}^\infty k\cdot \mathbb{P}\{X=k\} = \sum_{k=0}^\infty k \cdot e^{-\lambda}\frac{\lambda X^k}{k!}$$ $$=e^{-\lambda} \sum_{k=1}^\infty \frac{\lambda X^k}{(k-1)!} = e^{-\lambda}\cdot \lambda X \cdot \sum_{k=1}^\infty  \frac{\lambda X^{k-1}}{(k-1)!}$$ $$=\lambda Xe^{-\lambda} \sum_{k-1=0}^\infty \frac{\lambda X^{k-1}}{(k-1)!}$$ $$=\lambda X e^{-\lambda}e^{\lambda} \text{ since } e^X = \sum_{k=0}^\infty \frac{X^k}{k!}$$ thus $\mathbb{E}[X] = \lambda X,$ and similarly $\mathbb{E}[Y] = \lambda Y.$ Next, $$\mathbb{E}[X^2] = \sum_{k=0}^\infty k^2\cdot e^{-\lambda}\frac{\lambda X^k}{k!}$$ $$=\lambda_X^2 + \lambda_X.$$ $$\text{Var}[X] = M_2 - M_1^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \lambda X^2 + \lambda X - \lambda X^2 = \lambda X.$$ Similarly, $\text{Var}[Y] = \lambda Y.$
For (ii), $\mathbb{P}[X+Y = k] = \displaystyle{\sum_{i=0}^k \mathbb{P}(X+Y=k, Y=i)}$ $$=\sum_{i=0}^k \mathbb{P}(X=k-i) \mathbb{P}(Y=i)$$ $$=\sum_{i=0}^k \frac{e^{-\lambda x} \lambda X^{k-i}}{(k-i)!}\cdot \frac{e^{-\lambda Y}\lambda Y^i}{i!}$$ $$=\frac{1}{k!}\left[ \sum_{i=0}^k \frac{k!}{(k-i)!i!} \lambda X^{k-i} \lambda Y^i \right]e^{-(\lambda X + \lambda Y)}$$ $$=\frac{1}{k!}e^{-(\lambda X + \lambda Y)}\cdot (\lambda X + \lambda Y)^k.$$ So $X+Y$ is also a Poisson random variable with parameter $\lambda X + \lambda Y.$
For (iii) the characteristic function of Poisson random variable $X$ is $$M_X(Z) = \sum_{k=0}^\infty \mathbb{P}(X=k)\cdot e^{kZ}$$ $$=\sum_{k=0}^\infty e^{kZ} \frac{e^{-\lambda X}\lambda X^k}{k!}$$ $$=e^{-\lambda X} \sum_{k=0}^\infty \frac{(\lambda X e^Z)^k}{k!} = e^{-\lambda X} e^{e^Z\lambda X} = e^{\lambda X(e^Z - 1)}.$$ And $M_Y(Z) = e^{\lambda Y(e^Z-1)}.$ And $M_{X+Y}(Z)= M_X(Z) \cdot M_Y(Z)$ since $X$ and $Y$ are independent and equals $e^{\lambda X(e^Z - 1)}\cdot e^{\lambda Y(e^Z - 1)} = e^{(\lambda X + \lambda Y)e^Z-1}.$ Thus, $X+Y$ is also a Poisson random variable with parameter $\lambda X + \lambda Y.$
 A: For $n$ fixed we have $Z_{1,n}$, $Z_{2,n}$,$\dots$, $Z_{n,n}$ are iid Beornoulli randoms variables with paremeter $p=\frac{\lambda}{n}$, then we have that $Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n}$ have Binomial distribution where
\begin{align*} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}&= \binom{n}{k}p^{k}(1-p)^{n-k}\\
&=\binom{n}{k}\frac{\lambda^{k}}{n^{k}}\left(1-\frac{\lambda}{n}\right)^{n-k}\\
&=\binom{n}{k}\frac{\lambda^{k}(n-\lambda)^{n-k}}{n^{n}}.
\end{align*}
Therefore
\begin{align*}\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}&= \lim_{n\rightarrow\infty} \binom{n}{k}\frac{\lambda^{k}(n-\lambda)^{n-k}}{n^{n}}\\
&= \lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)\left(\frac{1}{k!}\right)\lambda^{k}\left(\frac{n-\lambda}{n}\right)^{n-k}\\
&=\lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)\left(\frac{1}{k!}\right)\frac{\lambda^{k}}{\left(\frac{n-\lambda}{n}\right)^{k}}\left(\frac{n-\lambda}{n}\right)^{n}\\
&=\lim_{n\rightarrow\infty} \color{red}{\left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)}\left(\frac{1}{k!}\right)\color{blue}{\left(\frac{\lambda n}{n-\lambda}\right)^{k}}\color{brown}{\left(1-\frac{\lambda}{n}\right)^{n}}\\
& \qquad\qquad \mbox{Note that }\\
& \qquad\qquad\quad \color{red}{\lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)=1,}\\
& \qquad\qquad\quad \color{blue}{\lim_{n\rightarrow\infty} \left(\frac{\lambda n}{n-\lambda}\right)^{k}=\lambda^{k}} \:\:\mbox{ and }\\
& \qquad\qquad\quad \color{brown}{\lim_{n\rightarrow\infty} \left(1-\frac{\lambda}{n}\right)^{n} =e^{-\lambda} }\\
&=\frac{\lambda^{k}}{k!}e^{-\lambda}.
\end{align*}
