Limit of $a_n=(1- \frac{1}{n})^n$ as $n\rightarrow \infty$ 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.
 A: 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
$$\left(1 - \frac{1}{n} \right)^n = \left(\frac{n-1}{n} \right)^n = \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}\left(1 + \frac{1}{n-1}\right)} \to \frac{1}{e \cdot 1} = e^{-1} $$
A: 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:
$$ \left( \left( 1+\frac{1}{n}\right)^{n+1} \right)^{-1} \rightarrow e^{-1}. $$
A: 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}. $
A: Simply write:
$$a_n= \left(1 - \frac{1}{n} \right)^n= \left(1 + \frac{(-1)}{n} \right)^{n } \rightarrow  e^{-1}=\frac{1}{e}$$
A: Or you could use
$$\lim_{n \to \infty}\bigg(1-\frac{1}{n}\bigg)^n = \lim_{n \to \infty}\bigg(1+\frac{1}{-n}\bigg)^{-n \cdot (-1)} = \bigg[\lim_{n \to \infty}\bigg(1+\frac{1}{-n}\bigg)^{-n}\bigg]^{-1} = e^{-1}$$
