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

Question

I need to prove $$\lim\limits_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = \frac{1}{e}$$ For now, I only know the e definition: $\lim\limits_{n\to\infty} \left ( 1+\frac{1}{n} \right)^n = e $, and I need to proof all the partial results in between to reach the wanted result.

Answer 1

Someone posted this in a comment before, but it appears to be deleted (too bad)

Observe, by first manipulating fractions, $$ \lim_{n\rightarrow\infty}\left(1-\frac{1}{n}\right)^{n} =\lim_{n\rightarrow\infty}\left(\frac{n-1}{n}\right)^{n} =\lim_{n\rightarrow\infty}\left(\frac{n}{n-1}\right)^{-n} =\lim_{n\rightarrow\infty}\left(\frac{n-1+1}{n-1}\right)^{-n} =\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{-n} $$ Then, by manipulating exponents, $$ =\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{-n(n-1)/(n-1)} =\lim_{n\rightarrow\infty}\left(\left(1+\frac{1}{n-1}\right)^{n-1}\right)^{-n/(n-1)} =\lim_{n\rightarrow\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^{-(n+1)/n}. $$ Then, the exponent goes to $-1$ and the inside goes to $e$, so the limit goes to $\frac{1}{e}$.

Answer 2

HINT:

For finite $r\ne0,m;$ $$\lim_{n\to\infty}\left(1+\dfrac rn\right)^{mn}=\left[\lim_{n\to\infty}\left(1+\dfrac rn\right)^{n/r}\right]^{mr}$$

Set $\dfrac nr= u$ and use the definition of $e$

Answer 3

Consider $$ \frac{1}{1-\dfrac{1}{n}}=\frac{n}{n-1}=\frac{n-1+1}{n-1}= 1+\frac{1}{n-1} $$ Then the limit of the reciprocal of your sequence is $$ \lim_{n\to\infty}\left(1+\frac{1}{n-1}\right)^{\!n}= \lim_{n\to\infty}\left(1+\frac{1}{n-1}\right)^{\!n-1} \left(1+\frac{1}{n-1}\right)=e\cdot 1=e $$

Answer 4

Use the definition of e and substitute $n$ by $-m$.

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

Therefore

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

Answer 5

$$1-\dfrac{1}{n}=\dfrac{n-1}{n}=\dfrac{1}{\dfrac{n}{n-1}}=\dfrac{1}{1+\dfrac{1}{n-1}}\tag{1}$$ by $(1)$ $$=\lim_{n\longrightarrow \infty}‎\Big(1-\dfrac{1}{n}\Big)^{n}$$ $$=\lim_{n\longrightarrow \infty}‎\Big(\dfrac{1}{1+\dfrac{1}{n-1}}\Big)^{n}$$ $$‎=\dfrac{1}{\lim_{n\longrightarrow \infty}\Big(1+\dfrac{1}{n-1}\Big)^{n}}$$ $$‎=\dfrac{1}{\lim_{n\longrightarrow \infty}\Big(1+\dfrac{1}{n-1}\Big)^{n-1}}$$ $$=\frac{1}{e}$$

Answer 6

There are obviously nicers answers on this page than this argument, but perhaps this adds something too. At any rate:

Write

$$ ( 1 - 1/n ) ^n = ( 1 - 1/n )^n \cdot { (1+1/n)^n \over (1+1/n)^n} = \def\l {( 1 - 1/n^2)^n}{\l \over ( 1 + 1/n )^n }.$$

We know that the denominator tends to $e$, so if the numerator (on the far right) tends to $1$, we are done.

Now, $$\l \le 1, $$ so the numerator is bounded above by $1$. We need to obtain a nice bound from below.

Expanding $\l$ using the binomial theorem, one has $$1 - {n \choose 1 }\def\r{\left(1\over n^2\right)}\r + { n \choose 2 }\r^2 - {n \choose 3} \r^3 + \cdots \ = \l$$ Dropping the $+$ terms on the left (except for the constant term $1$), one has

$$1 - {n \choose 1}\r - {n \choose 3} \r ^ 3 - \cdots \ \le \l.$$

Now, clearly $ {n \choose k } \le n^k$, so one gets from the previous line that
$$1 - n\r - n^3 \r ^ 3 - \cdots \ \le \l.$$ Canceling $n^k$ in the numerators and denominators, and taking out the common factor of $-1/n$, one gets

$$1 - {1\over n}\left\{ 1 + \r + \r^2 + \cdots \right\} \le \l.$$ For $n \ge 2$, the term in the brackets is $\le 4/3$ (geometric series), so $$ 1 - 4/(3n)\ \le\ \l \ \le\ 1.$$ Hence (by the 'sandwich theorem'), $\l$ tends to $1$, and the original sequence tends to $1/e$.