Limit of $a_n=(1- \frac{1}{n})^n$ as $n\rightarrow \infty$

Question

I want to calculate the limit of the following sequence: $$a_n=(1- \frac{1}{n})^n$$ First off I will calculate some terms to understand the behaviour: $$a_1=0 $$ $$a_2=\left(\frac{1}{2} \right)^2 =\frac{1}{4} $$ $$ \vdots$$ $$ a_{20}=\left(\frac{19}{20}\right)^{20} \approx 0.358$$ $$\vdots$$ $$ a_{100}=\left(\frac{99}{100}\right)^{100} \approx0.366$$ That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.

Answer 1

Using $(1 +x/n)^n \to e^x$ somewhat defeats the purpose of such a question.

If you are only given $(1 + 1/n)^n \to e$, then

$$\left(1 - \frac{1}{n} \right)^n = \left(\frac{n-1}{n} \right)^n = \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}\left(1 + \frac{1}{n-1}\right)} \to \frac{1}{e \cdot 1} = e^{-1} $$

Answer 2

Notice that: $$a_n=\left( \frac{n-1}{n}\right)^n =\left( \frac{n}{n-1}\right)^{-n}=\left( \frac{n-1+1}{n-1}\right)^{-n}= \left(\left( 1+\frac{1}{n-1}\right)^{n} \right) ^{-1}$$ For very large $n$, we have that by an index shift argument this will have the same limit as: $$ \left( \left( 1+\frac{1}{n}\right)^{n+1} \right)^{-1} $$ Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n \rightarrow \infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get: $$ \left( \left( 1+\frac{1}{n}\right)^{n+1} \right)^{-1} \rightarrow e^{-1}. $$

Answer 3

You can still write $a_n=\left(1-\dfrac{1}{n}\right)^n$ in terms of $b_n = \left(1+\dfrac{1}{n}\right)^n$. In fact, $a_n = \dfrac{b_n}{(b_{\frac{n-1}{2}})^{\frac{2n}{n-1}}}\to \dfrac{e}{e^2} = \dfrac{1}{e}. $

Answer 4

Simply write: $$a_n= \left(1 - \frac{1}{n} \right)^n= \left(1 + \frac{(-1)}{n} \right)^{n } \rightarrow e^{-1}=\frac{1}{e}$$

Answer 5

Or you could use

$$\lim_{n \to \infty}\bigg(1-\frac{1}{n}\bigg)^n = \lim_{n \to \infty}\bigg(1+\frac{1}{-n}\bigg)^{-n \cdot (-1)} = \bigg[\lim_{n \to \infty}\bigg(1+\frac{1}{-n}\bigg)^{-n}\bigg]^{-1} = e^{-1}$$