Asymptotic order of the convergence $\lim_{n\to\infty}\left(1-\frac{x}{n}\right)^n$

Question

For $x>0$ and $n\in\mathbb{N}$, we have $$ \lim_{n\to\infty}\left(1-\frac{x}{n}\right)^n=e^{-x}. $$ I would like to study the asymptotic order of the convergence. By a Taylor expansion, I think $$ \left(1-\frac{x}{n}\right)^n=e^{-x}+\mathcal{O}\left(\frac{x}{n}\right)^{2n}\;\mbox{as}\;n\to\infty. $$

Does anyone please give me some help or hints (or references)? Thanks so much!

Answer 1

We have that

$$\left(1-\frac{x}{n}\right)^n=e^{n\log\left(1-\frac{x}{n}\right)}$$

then use Taylor's expansion

• $\log(1-t)=t-\frac12t^2+O(t^3)$ as $t =\frac x n\to 0$

and therefore

$$e^{n\log\left(1-\frac{x}{n}\right)}=e^{n\left(-\frac xn-\frac12 \frac {x^2}{n^2}+O\left(\frac1{n^3}\right)\right)}=e^{-x}\cdot e^{-\frac12 \frac {x^2}{n}+O\left(\frac1{n^2}\right)}=e^{-x}\left(1-\frac12 \frac {x^2}{n}+O\left(\frac1{n^2}\right)\right)$$

that is

$$\left(1-\frac{x}{n}\right)^n=e^{-x}-e^{-x}\frac12 \frac {x^2}{n}+O\left(\frac1{n^2}\right)$$

Answer 2

Consider $$a_n=\left(1-\frac{x}{n}\right)^n\implies \log(a_n)=n \log\left(1-\frac{x}{n}\right)$$ Using Taylor expansion $$\log(a_n)=n\left(-\frac{x}{n}-\frac{x^2}{2 n^2}-\frac{x^3}{3n^3}+O\left(\frac{1}{n^4}\right)\right)=-x-\frac{x^2}{2 n}-\frac{x^3}{3n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor $$a_n=e^{\log(a_n)}=e^{-x}-e^{-x}\frac{ x^2}{2 n}+e^{-x}\frac{ x^3 (3 x-8)}{24 n^2}+\cdots$$