triangular arrays in law of rare events theorem In Durrett's probability theory and examples,

Theorem 3.6.1 For each $n$ let $X_{n,m}, 1 \le m \le n$ be independent random variables with $P(X_{n,m} = 1) = p_{n,m}, P(X_{n,m} = 0) = 1- P_{n,m}$. Suppose
(i) $\sum_{m=1}^n p_{n,m} \to \lambda \in (0, \infty)$
and (ii) $\max_{1 \le m \le n} p_{n,m} \to 0$.
If $S_n = X_{n,1} + \cdots + X_{n,n}$ then $S_n \Rightarrow Z$ where $Z$ is $Poisson(\lambda)$.

I don't understand the why we use triangular arrays. For instance, why can't we let $X_n$ be independent random variables such that $P(X_n = 1) = p_n, P(X_n = 0) =1-p_n$. Condition (i) would be $\sum_{n=1}^K p_n \to \lambda$ as $K \to \infty$ and $p_{n} \to 0$ as $n \to \infty$. Then $X_n \Rightarrow Z$.
 A: In the setting you suggest, we do have that $\sum_{n=1}^NX_n\to Y$ in distribution to some random variable $Y$, which takes integer values. Moreover, we  have to take $p_n\neq 1$ for each $n$, otherwise, the event $\{Y=0\}$ would have probability zero.
We have
$$
\mathbb P(Y=0)=\prod_{n=1}^\infty (1-p_n), \quad \mathbb P(Y=1)=\sum_{i\geqslant 1}p_i\prod_{n=1,n\neq i}^\infty (1-p_n).
$$
If $Y$ had a Poisson distribution, then we should have
$$
e^{-\lambda}=\prod_{n=1}^\infty (1-p_n), \quad e^{-\lambda}\lambda=\sum_{i\geqslant 1}p_i\prod_{n=1,n\neq i}^\infty (1-p_n).
$$
Since $$
\prod_{n=1,n\neq i}^\infty (1-p_n)=\frac1{1-p_i}\prod_{n=1 }^\infty (1-p_n),
$$
we get
$$
e^{-\lambda}=\prod_{n=1}^\infty (1-p_n), \quad e^{-\lambda}\lambda=\sum_{i\geqslant 1}\frac{p_i}{1-p_i}e^{-\lambda}.
$$
Taking the logarithm in the first equality gives
$$
\lambda=\sum_{n=1}^\infty\ln\left(\frac 1{1-p_n}-1+1\right)=\sum_{i\geqslant 1}\frac{p_i}{1-p_i} 
$$
and since the inequality $\ln(1+x)\leqslant x$ holds only for $x=0$, we get a contradiction.
