Motivated by Poisson distributions, prove that: If $n p_n \to \lambda$ then $ \left( 1 - p_n \right)^n \to e^{-\lambda}$ I have this ugly proof that some sequence converges to something and I really don't like it because it seems too hard for no reason... can anyone help me make this more simple?
Here it is : We are given a sequence $p_n$ with $n p_n \to \lambda > 0$. We show that 
$$
\left( 1 - p_n \right)^n \to e^{-\lambda}.
$$
So my proof begins by noticing that
$$
(1-p_n)^n = \left( 1 + \frac{-np_n}n \right)^n 
$$
and that the family of functions $s_n(x) = \left(1+\frac xn \right)^n$ is equicontinuous at $x$ (I think I can pull some argument stating that the derivative of $s_n(x)$ is always bounded (because it also converges to $e^x$) when considering an interval of arbitrary but fixed length centered at $x$, so that in this interval, all functions $s_n$ can have the same Lipschitz constant, therefore giving equicontinuity, correct me if I'm wrong) so that if I note by $x_n$ the sequence $-n p_n$, $x = -\lambda$ and $s(x) = e^x$, I wish to show that 
$$
\forall \varepsilon > 0, \quad \exists N \quad s.t. \quad \forall n > N, \quad |s_n(x_n) - s(x)| < \varepsilon.
$$
so now,
$$
|s_n(x_n) - s(x)| \le |s_n(x_n) - s_n(x_m)| + |s_n(x_m) - s_n(x)| + |s_n(x) - s(x)|.
$$
Since $x_n$ is convergent, it is also Cauchy, so that for all $\delta > 0$, there exists an $N_1$ such that for all $n,m > N_1$, $|x_m - x_n| < \delta$ and $|x_m - x| < \delta$. Since the family $\{s_n\}$ is equicontinuous, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that $|x - y| < \delta \Rightarrow |s_n(x) - s_n(y)| < \frac{\varepsilon}3$, so that the first two terms are taken care of by $N_1$ by taking $(x,y) = (x_n, x_m)$ and $(x,y) = (x_m, x)$ respectively. The last term is also bounded by $\varepsilon/3$ when $n > N_2$ because $s_n \to s$ pointwisely. Taking $N = \max \{ N_1, N_2 \}$ we're done.
I feel like my proof is a little too harsh on this easy-looking problem... anyone has a better proof? What I mean by "better" is using less powerful or advanced tools, something simple. I can't seem to find anything better right now.
 A: This could even be done by a(n admittedly very strong) calculus student.
Step 1: For any real number $l$, one has $\lim_{n \rightarrow \infty} (1-\frac{l}{n})^n = e^{-l}$.  (Take logarithms and apply L'Hopital's Rule.)
Step 2: Let $M > \lambda$, so that for sufficiently large $n$, $(1-p_n)^n \geq (1-\frac{M}{n})^n$.  By Step 1, for any $\epsilon > 0$ and all sufficiently large $n$ we have 
$(1-p_n)^n \geq e^{-M} - \epsilon$.
Step 3: Similarly, let $m < \lambda$ and arguing as above for any $\epsilon > 0$ and all sufficiently large $n$ we have 
$(1-p_n)^n \leq e^{-m} + \epsilon$.
The result follows!
A: It is also quite natural to attack the $n$th power
directly using the binomial theorem:
$$\begin{align*}(1-p_n)^n&=\sum_{j\geq 0}{n\choose j}(-p_n)^j\\\\&=
\sum_{j\geq 0}\left[n(n-1)\cdots(n-j+1)\over n^j\right](-np_n)^j/j!\\\\
&\to \sum_{j\geq 0}(-\lambda)^j/j!\\\\&=\exp(-\lambda).\end{align*}
$$
The convergence is obvious termwise, but you need the Dominated 
Convergence Theorem (say), to justify convergence of the infinite sums. 
A: Hmmmm... This smells like plain analysis to me: you want to show that $n\log(1-p_n)\to-\lambda$. 
First, $\log(1-x)\le -x$ for every $x$ in $(0,1)$ hence $n\log(1-p_n)\le-np_n$. Since $np_n\to\lambda$, $\limsup n\log(1-p_n)\le-\lambda$. 
Second, $\log(1-x)\ge-x-x^2$ for every $x$ in $(0,\frac12)$. For every $n$ large enough, $p_n\le\frac12$ hence $n\log(1-p_n)\ge-np_n-np_n^2$. Since $np_n^2\to0$, $\liminf n\log(1-p_n)\ge-\lambda$. 
You are done.
