$(\star)$ : Let $ X_n $ ~ Bin($n,p_n$) ( $ n \in \mathbb{N} $ )
$ n \cdot p_n \rightarrow \lambda $ for $ n \rightarrow \infty $.
Then $ X_n$ convergences to Poisson $Z$ ~ Poisson($\lambda$).
I have to use the following steps:
a) First of all solve this exercise for $ n \cdot p_n = \lambda $ $ \forall n \in \mathbb{N}$.
b) Let $[A,B]$ be a bounded interval. Show $ \forall k \in \mathbb{N}$ $ \forall a,b \in [A,B]$:
$| a^k - b^k | = | \int_{min(a,b)}^{max(a,b)} k \cdot x^{k-1} dx | \le k \cdot| a-b| \cdot $ max{ |A|,|B| }$^{k-1}$
c) Let $[A,B]$ be a bounded interval. Show there is a $D > 0 $, such that $\forall a,b \in [A,B]$ $ \forall n \in \mathbb{N} $:
| $(1+\frac{a}{n})^n - (1+\frac{b}{n})^n | \le D|a-b| $.
d) Now proof ($\star$).
Attempt:
a) $n \cdot p_n = \lambda \Rightarrow p_n = \frac{\lambda}{n}$. So: $\lim_{n \rightarrow \infty} P(X_n = k) = \lim_{n \rightarrow \infty} \binom{n}{k} p_n^k(1-p_n)^{n-k} =$ $\lim_{n \rightarrow \infty} \binom{n}{k} (\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k} = \frac{\lambda^k}{k!} \lim_{n \rightarrow \infty} \frac{n}{n} \frac{n-1}{n} ... \frac{n-k+1}{n}(1-\frac{\lambda}{n})^n \frac{1}{(1-\frac{\lambda}{n})^k} = \frac{\lambda^k}{k!} \cdot 1\cdot e^{-\lambda} \cdot 1 = e^{-\lambda} \frac{\lambda^k}{k!} $.
b) Without loss of generality: $ a \le b$.
$| \int_{min(a,b)}^{max(a,b)} k \cdot x^{k-1} dx | = | \int_{a}^{b} k \cdot x^{k-1} dx | = | x^k|_{a}^{b} | = |b^k -a^k| = |a^k - b^k| $.
second part: for $a = b$ we have $ 0 \le 0 $ and for $ a < b $ we use the mean-value theorem:
$\frac{|a^k - b^k|}{|a-b|} = k \cdot |c|^{k-1} \le k \cdot $ max{|A|,|B|}$^{k-1}$ because $c \in (a,b) \subset [A,B]$.
c) Again the mean-value theorem? $| \frac{(1+\frac{a}{n})^n - (1+\frac{b}{n})^n}{a-b} | = |(1+\frac{c}{n})^{n-1}| $ but why is this $ \le D $ ? I know that $c \in (a,b) $.
d) So now I have to show that for random variables $X_n$ ~ Bin$(n,p_n)$ with $n \cdot p_n \rightarrow \lambda$ for $ n \rightarrow \infty $ $\Rightarrow $ $X_n $ convergences to $Z$ ~ Poisson($\lambda$). Here is my problem. Unfortunately I don't know how I can use a), b) and c) for this exercise. I hope that your answer can help me.
Important remark: It is allowed to use:
Let $X_n$ be random variables with values in $\mathbb{N_0}$. If the sequence of functions $g_n(t) := g_{X_n}(t)$ convergences pointwise to $g(t)$ on a open interval $I$ with $0 \in I $, then $\exists$ a random variable $X$ with $g(t) = g_X(t) $ and $ \lim_{ n \rightarrow \infty } P (X_n = k) = P(X=k)$ $ \forall k \in \mathbb{N_0}$.
It is much easier to approach this through the use of generating functions. Let $X_i$ be an i.i.d. sequence of $\mathsf{Ber}(p)$ distributed random variables for nonnegative integers $i$, then $X_1$ has probability generating function $P(s) = 1 - (1-z)p$, and so $S_N:=\sum_{i=0}^N X_i$ has probability generating function $$ P_N(s;p) = P(s)^N = (1+(s-1)p)^N. $$ Taking the limit as $N\to\infty$ while keeping $pN\equiv \lambda$ constant yields $$ \lim_{N\to\infty} P_N(s;p) =\lim_{N\to\infty} \left(1+\frac{(s-1)\lambda}N\right)^N = e^{(s-1)\lambda} = \sum_{i=0}^\infty \frac{e^{-\lambda}\lambda^k}{k!}s^k, $$ which is the probability generating function for the $\mathsf{Pois}(\lambda)$ distribution.