Approximation of Binomial distribution with Poisson I have to prove $\sup|P_{n}(k)-\pi_{\lambda}(k)| \leq \frac{\lambda^2}{n}$ where $P_{n}(k)$ is distribution of Binomial RV with parameters $(n, \frac{\lambda}{n})$ and $\pi_{\lambda}(k)$ is Poisson with parameter $\lambda$. 
I've shown that if $\xi=\sum_{j=1}^{n}\xi_j$, where $\xi_j \sim \text{Ber} \left(\frac{\lambda}{n}\right)$ and $\eta=\sum_{j=1}^{n} \eta_j$, where $\eta_j \sim \text{Poi} \left(\frac{\lambda}{n}\right)$, then $\xi$ has probability distribution $P_{n}(k)$ and $\eta \ \pi_{\lambda}(k)$.
Any suggestions will be massively appreciated. 
 A: See pages 20-21 of the book Stochastic processes and models, by David Stirzaker, here.
Adapting the linked solution to your case and notation: Using your notation, choose a joint distribution $f(x,y)$ for $\xi_j$ and $\eta_j$ as follows:
$$
f(0,0) = 1 - \lambda /n,
$$
$$
f(1,0) = e^{ - \lambda /n}  - 1 + \lambda /n,
$$
$$
f(1,y) = \frac{{(\lambda /n)^y e^{ - \lambda /n} }}{{y!}}, \;\; y \geq 1,
$$
$$
f(0,y) = 0,\;\; y \geq 1.
$$
Note that $f(\cdot,\cdot)$ indeed determines a joint distribution, as
$$
f(0,0) + f(1,0) + \sum\limits_{y = 1}^\infty  {f(1,y)}  + f(0,y) = 1.
$$
Further note that $\xi_j$ and $\eta_j$ are not independent with this joint distribution. It holds
$$
{\rm P}(\xi _j  \ne \eta _j ) = f(1,0) + {\rm P}(\eta _j  \ge 2) = \lambda /n - (\lambda /n)e^{ - \lambda /n}  \le (\lambda /n)^2 .
$$
Next, using your notation,
$$
|{\rm P}(\xi  = k) - {\rm P}(\eta  = k)| \le {\rm P}(\xi  \ne \eta ) \le {\rm P}(\xi _j  \ne \eta _j ,\;{\rm for}\;{\rm some}\; j)
$$
(the second inequality is trivial; the first is derived in the edit below).
Thus,
$$
|{\rm P}(\xi  = k) - {\rm P}(\eta  = k)| \leq \sum\limits_{j = 1}^n {{\rm P}(\xi _j  \ne \eta _j )}  \leq \frac{{\lambda ^2 }}{n},
$$
and so
$$
|P_n (k) - \pi _\lambda  (k)| \le \frac{{\lambda ^2 }}{n}.
$$
EDIT: Derivation of the inequality
$$
|{\rm P}(\xi  = k) - {\rm P}(\eta  = k)| \le {\rm P}(\xi  \ne \eta ) .
$$
From
$$
{\rm P}(\xi  = k) - {\rm P}(\eta  = k) = {\rm P}(\xi  = k,\eta  \ne k) + {\rm P}(\xi  = k,\eta  = k) - {\rm P}(\eta  = k),
$$
it follows that
$$
{\rm P}(\xi  = k) - {\rm P}(\eta  = k) \leq {\rm P}(\xi  = k,\eta  \ne k) \leq {\rm P}(\xi \neq \eta).
$$
Analogously, ${\rm P}(\eta  = k) - {\rm P}(\xi  = k) \leq {\rm P}(\xi \neq \eta)$; hence the result.
Also, in view of the proof (above the edit), it is worth noting that
$$
{\rm P}(\xi _j  \ne \eta _j ,\;{\rm for}\;{\rm some}\; j) = {\rm P}(\xi_1 \neq \eta_1  \vee \cdots \vee \xi_n \neq \eta_n) \leq \sum\limits_{j = 1}^n {{\rm P}(\xi _j  \ne \eta _j )} .
$$
