Approximation: (1 - 1/n) ^ m ~~ e^ (-m/n), why?

Question

$\left(1 - \frac{1}{n}\right) ^ m \approx e^{-m/n}$

Can someone explain why the left hand side is approximated by $e ^{-m/n}$ ?

Answer 1

As pointed out by one of the comments, there is no indeterminate form. In fact the sequence converges to $1$. If $m$ depends of $n$ or the expression is part of a bigger term, it must be considered as a whole.

Use $\lim_{x\rightarrow0}(1+x)^\frac{1}{x}=e$

In your case $x=-\frac{1}{n}\rightarrow 0$ so:

$\require{enclose}\enclose{horizontalstrike}{\lim_{n\rightarrow\infty}(1-\frac{1}{n})^m=\lim_{n\rightarrow\infty}((1-\frac{1}{n})^{-n})^{-\frac{m}{n}}\rightarrow e^{-\frac{m}{n}}}$

$\lim_{n\rightarrow\infty}(1-\frac{1}{n})^m=\lim_{n\rightarrow\infty}((1-\frac{1}{n})^{-n})^{-\frac{m}{n}}\rightarrow e^0=1$

Answer 2

Probaly too simplistic.

Consider $$A=\left(1-\frac{1}{n}\right)^{m}\implies \log(A)=m\log\left(1-\frac{1}{n}\right)$$

Now, using Taylor series for large $n$ $$\log\left(1-\frac{1}{n}\right)=-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ $$\log(A)=-\frac{m}{n}+O\left(\frac{1}{n^2}\right)$$ $$A=e^{\log(A)}=e^{-\frac{m}{n}+O\left(\frac{1}{n^2}\right)}\approx e^{-\frac{m}{n}}$$

Using one extra term $$\log\left(1-\frac{1}{n}\right)=-\frac{1}{n}+O\left(\frac{1}{n^2}\right)-\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A)=-\frac{m}{n}-\frac{m}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$A=e^{-\frac{m}{n}}\times e^{-\frac{m}{2n^2}+O\left(\frac{1}{n^3}\right)}$$