Show that for large $n$, $(1-\frac{\lambda}{n})^n$ is approximately $exp(-\lambda)$ In their book on Probability, Grinstead a Snell say (page 189) that for large $n$,  $(1-\frac{\lambda}{n})^n $ is approximately equal to $exp(-\lambda)$
Using the binomial formula I can expand $(1-\frac{\lambda}{n})^n $  to be $\sum^{n}_{j=0}\frac{(n)_j}{n^j}\cdot \frac{\lambda^j}{j!} $ so the sum is less than  $exp(-\lambda)$. But I don't see how as $n$ gets large its "approximately" the same. 
 A: Here is a proof that does not use the binomial formula, and reveals a bound on the error of the approximation for a given $n>x$:
$$0 \leqslant \left|e^{-x} - \left(1- \frac{x}{n}\right)^n\right| \leqslant e^{-x}\frac{x^2}{n}$$
Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} <  \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x < \left(1 - \frac{x}{n}\right)^{-n}.$$
Hence, using the Bernoulli inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n,$
$$0 \leqslant e^{-x} - \left(1- \frac{x}{n}\right)^n = e^{-x}\left[1 - e^x\left(1- \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{-x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{-x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{-x}\frac{x^2}{n}.$$
Therefore, 
$$0 \leqslant\sup_{x \in [0,\infty)} \left|e^{-x} - \left(1- \frac{x}{n}\right)^n\right| \leqslant \sup_{x \in [0,\infty)}e^{-x}\frac{x^2}{n}= \frac{4e^{-2}}{n}\xrightarrow[n \rightarrow \infty]\quad 0.$$
A: Use L'hospital's rule to prove that $$\lim_{n \to \infty}\left(1+\frac{c}{n}\right)^n=e^c$$ This is an exercise that many first year calculus students will encounter (or the more specific $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e$). But once you have established this limit, you can substitute $c = -\lambda$. It should stand to reason from here that $\left(1+\frac{c}{n}\right)^n\approx e^c$ for large enough $n$. For a more rigorous proof, you could argue that $$\left\{\left(1-\frac{\lambda}{n}\right)^n\right\}_{n=1}^\infty$$ is a strictly increasing sequence. You already know it's bounded above by $e^{-\lambda}$, so for all $\varepsilon>0$ there exists some $N \in \Bbb{N}$ such that $\left(1-\frac{\lambda}{n}\right)^n$ will be within $\varepsilon$ away from $e^{-\lambda}$ for all $n\geq N$. Let $\varepsilon$ represent how close of an approximation you want.
A: We have
$$\lim_{n\to \infty} \left(1 - \frac{\lambda}{n}\right)^n 
  = \left[\lim_{n\to \infty} \left(1 - \frac{\lambda}{n}\right)^{-n/\lambda}\right]^{-\lambda }
  = \left[\lim_{n\to \infty} \left(1 + \frac{1}{n}\right)^n\right]^{-\lambda}
  = e^{-\lambda}.
$$
