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.

The original problem is another one, and this is the step where i am stucked. — Ygdra, Jul 05 '15 at 16:12
I tryed appliyng a substitution but that seems to not work.. — Ygdra, Jul 05 '15 at 16:12
Do you know that $\lim\limits_{n\to-\infty} \left ( 1+\frac{1}{n} \right)^n = e$ — Eclipse Sun, Jul 05 '15 at 16:14
Yes. but the only way I know to make that minus appears is using substitutuin, and I don't know how that help me. — Ygdra, Jul 05 '15 at 16:15
See also: http://math.stackexchange.com/questions/358830/about-lim-left1-frac-xn-rightn and http://math.stackexchange.com/questions/596771/finding-the-limit-of-left-fracnn1-rightn — Martin Sleziak, Jul 05 '15 at 17:30
Can you show the product $(1+1/n)^n \cdot (1-1/n)^n$ goes to $1$? — GEdgar, Jul 05 '15 at 18:11

score 3 · Accepted Answer · answered Jul 05 '15 at 16:23

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 $\frac{1}{e}$.

score 3 · Answer 2 · answered Jul 05 '15 at 16:24

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

score 3 · Answer 3 · answered Jul 05 '15 at 16:29

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

score 1 · Answer 4 · answered Jul 05 '15 at 16:27

1

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

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

answered Jul 05 '15 at 16:27

callculus42

30,550

You're implicitly using what you want to prove. – egreg Jul 05 '15 at 16:29
I used the definition of e. I thought, that is what the OP wanted. – callculus42 Jul 05 '15 at 16:30
Well, the passage from $-m\to-\infty$ to $m\to\infty$ (you miss changing a minus into plus, by the way) has no justification. – egreg Jul 05 '15 at 16:32
If -m goes to $-\infty$, then m goes to $\infty$ – callculus42 Jul 05 '15 at 16:34
Here one of the answers have done the same : http://math.stackexchange.com/questions/524957/show-that-lim-x-to-infty-left-1-frac-lambdax-rightx-e-lambda?rq=1 – callculus42 Jul 05 '15 at 16:37
It's a very different setting: functions versus sequences. – egreg Jul 05 '15 at 16:39
Ok. Maybe you should first leave a comment there. It has 4 upvotes and a checkmark. I don´t see any different setting. – callculus42 Jul 05 '15 at 16:42

score 0 · Answer 5 · answered Jul 05 '15 at 16:14

0

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 $e$

answered Jul 05 '15 at 16:14

lab bhattacharjee

274,582

In any case you're using a much general result, which is $\lim_{x\to\infty}(1+(1/x))^x=\lim_{x\to-\infty}(1+(1/x))^x=e$ (where $x$ is used to emphasize that this is not the limit of a sequence). – egreg Jul 05 '15 at 16:26

score 0 · Answer 6 · answered Jul 06 '15 at 03:10

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 $1/e$.

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

6 Answers6

Linked