# limit of $\left( 1-\frac{1}{n}\right)^{n}$

limit of $$\left( 1-\frac{1}{n}\right)^{n}$$

is said to be $\frac{1}{e}$ but how do we actually prove it?

I'm trying to use squeeze theorem

$$\frac{1}{e}=\lim\limits_{n\to \infty}\left(1-\frac{1}{n+1}\right)^{n}>\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n} > ??$$

• What is your definition of $e$? – anomaly Nov 16 '15 at 4:50
• I think he uses $lim_{n\rightarrow \infty}(1+{1\over n})^n$ – cr001 Nov 16 '15 at 4:52
• Your inequality is wrong. Just because one sequence is strictly greater than another at every term does not mean that its limit (if it exists) is strictly greater than the limit of the other (if it exists). – user21820 Nov 16 '15 at 5:04
• See also math.stackexchange.com/questions/596771/… and maybe some posts linked there. – Martin Sleziak Nov 16 '15 at 7:59
• You might find some related posts also among frequent questions tagged limits+e. – Martin Sleziak Nov 16 '15 at 8:00

It easily follows from $(1+\frac1n)^n \longrightarrow e$.

Infact, $(1-\frac1n)^n=(\frac{n-1}{n})^n=(\frac{n}{n-1})^{-n}=(1+\frac1{n-1})^{-n} \longrightarrow e^{-1}$

• How to rigorously show the last step? – gabrielchua Nov 16 '15 at 4:58
• @gabrielchua: Why don't you try yourself by manipulating the last expression before $e^{-1}$ into something that looks like the definition of $e$? – user21820 Nov 16 '15 at 5:01
• Anyway, Ottavio, it is best to specify how the limit is taken whenever you use "$\to$", in this case "as $n \to \infty$". – user21820 Nov 16 '15 at 5:05
• @user21820 I took it for granted, since $+\infty$ is the only accumulation point for $\mathbb{N}$ – Ottavio Bartenor Nov 16 '15 at 5:10
• Well you didn't specify what $n$ is either! Technically $n$ is just some arbitrary variable, and nothing to do with $\mathbb{N}$. – user21820 Nov 16 '15 at 5:11

$$\left(1-\frac{1}{n+1}\right)^{n}>\left( 1-\frac{1}{n} \right)^{n}>\left( 1-\frac{1}{n} \right)^{n+1}$$

and

$$\lim\limits_{n\to \infty}\left(1-\frac{1}{n+1}\right)^{n}={1\over e}$$

$$\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n+1}=\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n-1}\cdot\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{2}=\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n-1} \cdot 1$$

$$= \lim\limits_{n-1\to \infty} \frac{1}{ \left( 1+\frac{1}{n-1} \right)^{n-1} } = \frac{1}{ \lim\limits_{n-1\to \infty} \left( 1+\frac{1}{n-1} \right)^{n-1} } = {1\over e}$$

Hence by squeeze theorem you know

$$\lim\limits_{n\to \infty}\left(1-\frac{1}{n}\right)^{n}={1\over e}$$

As discussed in the comment, your way of writing is not formally correct and you cannot use the squeeze theorem in one line.

• Your first and second inequalities are wrong. – user21820 Nov 16 '15 at 5:02
• It comes from $1>(1-{1\over n})$. I know I should not put in the limit notation but I am following the OP's way of writing. (to use the sandwich theorem) – cr001 Nov 16 '15 at 5:04
• But you should not follow the asker's way of writing if it is mathematically incorrect, but rather you should point out the error if there is one. – user21820 Nov 16 '15 at 5:06
• Yep you are right, I will edit the answer and point out the error. – cr001 Nov 16 '15 at 5:07
• See if you're happy with my edit to add in one step. – user21820 Nov 16 '15 at 5:53

Alternately $$e^x=\lim \left(1+\frac{x}{n}\right)^n$$ put $x=-1$

• This doesn't address the spirit of the question. – Mark Viola Nov 16 '15 at 5:58

Another way considering $$A=\left( 1-\frac{1}{n}\right)^{n}$$ Using logarithms $$\log(A)=n\log\left( 1-\frac{1}{n}\right)$$ Now, using Taylor expansion for small values of $x$ $$\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$$ replacing $x$ by $\frac 1n$ $$\log(A)=n\left(-\frac 1 n -\frac{1}{2n^2}+\cdots\right)=-1-\frac{1}{2n}+\cdots$$

I am sure that you can take from here.

I believe you just want the proof. I am not using squeeze theorem in this answer because you haven't said that you want to prove it only using squeeze theorem. This is a limit of the form $1^{\infty}$ which can be easily done by using the following steps( Here $\lim_{x \to 0}f(x)=0$ and $\lim_{x \to 0}g(x)=\infty$):-\begin{align}\lim_{x \to 0}(1+f(x))^{g(x)}&=\lim_{x \to 0}e^{g(x)\ln[1+f(x)]}\\&=e^{\lim_{x \to 0}g(x)[1+f(x)-1]}\\&=e^{\lim_{x \to 0}g(x)f(x)}\end{align}Here, I have used the fact that $\lim_{h(x) \to 1}\ln(h(x))=h(x)-1$. In most of these questions, it is easy to find $\lim_{x \to 0}g(x)f(x)$ which is of the form $\infty*0$. For your question $f(x)=-x$ and $g(x)=\frac 1x$ (In your limit $n \to \infty$ and in my limit $x\to0$, hence the difference). So $\lim_{x\to0}g(x)f(x)=-1$. Therefore your limit is $e^{-1}=\frac 1e$.

If one wants to use the squeeze theorem, then we can proceed as follows.

First, We define

$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n \tag 1$$

Then, we note that

$$\left(1-\frac1n\right)\left(1+\frac1n\right)=\left(1-\frac{1}{n^2}\right)<1\implies \left(1-\frac1n\right)^n <\frac{1}{\left(1+\frac1n\right)^n } \tag 2$$

Also, using Bernoulli's Theorem we have

$$\left(1-\frac{1}{n^2}\right)^n\ge 1-\frac1n\implies \left(1-\frac1n\right)^n \ge \frac{1-\frac1n}{\left(1+\frac1n\right)^n} \tag 3$$

Putting $(2)$ and $(3)$ together reveals that

$$\frac{1-\frac1n}{\left(1+\frac1n\right)^n} \le \left(1-\frac1n\right)^n \le \frac{1}{\left(1+\frac1n\right)^n }$$

whence taking the limit as $n\to \infty$, using $(1)$ along with the squeeze theorem provides the anticipated result

$$\lim_{n\to \infty}\left(1-\frac1n\right)^n=e^{-1}$$

• Please let me know how I can improve my answer. I really want to give the best answer I can and this one addresses your request to use the squeeze theorem directly. – Mark Viola Nov 22 '15 at 6:02