A question regarding $e^{-\lambda}\approx \left(1-\frac{\lambda}{n}\right)^n$

Question

For large $n$ and moderate $\lambda$, $e^{-\lambda}\approx \left(1-\frac{\lambda}{n}\right)^n$. Moderate $\lambda$ however does not say very much, is there some way to be a bit more specific? Is there perhaps some term that shows the remainder from the actual value of $e^{-\lambda}$ that can be used to show what moderate $\lambda$ ought to mean?

Answer 1

One possible way to get error bounds is to use logarithms on both sides. Note that for example $$\ln \left(\left(1 - \frac{\lambda}{n}\right)^n\right) = n \ln \left(1 - \frac{\lambda}{n}\right) = n \left(-\frac{\lambda}{n} + O\left(\frac{\lambda^2}{n^2}\right)\right) = -\lambda + O\left(\frac{\lambda^2}{n}\right),$$

which yields

$$e^{-\lambda} = \left(1 - \frac{\lambda}{n}\right)^n \exp\left(-O\left(\frac{\lambda^2}{n}\right)\right).$$

If you want some more explicit bounds, then you need to analyze the remainder of the Taylor expansion of the logarithm more precisely.

For example, the following two bounds hold for all $x > 0$: $$1 - \frac{1}{x} \le \ln(x) \le x - 1$$

(See here and here). This implies that for $x < 1$ we have $$\frac{-x}{1 - x} \le \ln(1 - x) \le -x,$$ which gives us the following bounds in the case $\lambda < n$: $$\exp\left(-\frac{\lambda}{1 - \frac{\lambda}{n}}\right) \le \left(1 - \frac{\lambda}{n}\right)^n \le e^{-\lambda}.$$

Answer 2

Take the case of $n=3$. By the Binomial theorem,

$$\left(1-\frac\lambda 3\right)^3=1-\lambda+\frac{\lambda^2}3-\frac{\lambda^3}{27},$$

to be compared to

$$1-\lambda+\frac{\lambda^2}2-\frac{\lambda^3}{3!}+\frac{\lambda^4}{4!}\cdots$$

More generally, the coefficient of the quadratic term is

$$\frac1{n^2}\binom n2=\frac{n-1}{2n},$$ which never reaches $1/2$. So a first approximation of the truncation error is

$$\frac{\lambda^2}{2n}.$$