3
$\begingroup$

I would like to show $\lim_{n \rightarrow \infty}\left(\frac{n - 1}{n}\right)^n = 1/e$.

I know the argument typically goes like this: Let $y = \left(\frac{n - 1}{n}\right)^n$. Then $\ln(y) = n\cdot{}\ln \left(\frac{n - 1}{n}\right)$. Taking the limit as $n \rightarrow \infty$, we have an indeterminant product of the form $\infty\cdot0$.

I think ideally I would like to use L'Hopital's Rule, so the issue is getting this into the correct form to apply it. I don't think the simplification $n\ln(n - 1) - n\ln(n)$ helps any.

But if we can establish that $\lim_{n\rightarrow\infty}\ln(y) = -1$, then using the identity $y = e^{\ln(y)}$, we'd arrive at the desired result.

Alternatively, could one use the definition of $e$? This might not help, but $e = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = \lim_{n \rightarrow \infty}(\frac{n + 1}{n})^n$, which looks similar to what we have, but not quite.

$\endgroup$
2
$\begingroup$

The idea is good; we try proving that $$ \lim_{x\to\infty}\left(\frac{x-1}{x}\right)^x $$ exists; if it does, then it's the same as the limit of your sequence. So we try finding the limit of the logarithm: $$ \lim_{x\to\infty}x\log\frac{x-1}{x} $$ Now do the substitution $x=1/t$, which brings the limit to the form $$ \lim_{t\to0^+}\frac{1}{t}\log(1-t) $$ which is the derivative at $0$ of the function $f(t)=\log(1-t)$.

Why doing $t=1/x$? For two reasons. First $$ \frac{x-1}{x}=1-\frac{1}{x}=1-t $$ Second, we have $x$ as a factor, which becomes $t$ at the denominator. Third, this transforms a limit at $\infty$ to a limit at $0$, where derivatives are defined.

$\endgroup$
1
$\begingroup$

Consider making the change of variable $k=n-1$ here to get $$\lim_{n\rightarrow\infty}\left(\frac{n-1}{n}\right)^n = \lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k+1} =\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k} =\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k}$$

which looks like the reciprocal of the limit definition of $e$.

If you want to use your original approach you can write $\ln(y)$ as $$\lim_{n\rightarrow\infty}\frac{\ln(\frac{n-1}{n})}{\frac{1}{n}}$$ which has indeterminate form $\frac{0}{0}$ so we use L'Hopital's to get this as $$\lim_{n\rightarrow\infty}\frac{\frac{n}{n-1}\frac{1}{n^2}}{\frac{-1}{n^2}} = -\lim_{n\rightarrow\infty}\frac{n}{n-1 } = -1 $$

$\endgroup$
  • $\begingroup$ It's better to write $\log(n-1)-\log n$ at the numerator, before computing the derivative, which is $\frac{1}{n-1}-\frac{1}{n}$. $\endgroup$ – egreg Oct 13 '14 at 20:10
1
$\begingroup$

You can use the definition $e = \lim_{n \rightarrow \infty} (1+1/n)^n$.

Note that

$$\lim\left(\frac{n-1}{n}\right)^{-n}= \lim\left[\left(1+\frac{1}{n-1}\right)^{n-1}\left(1+\frac{1}{n-1}\right)\right]\\= \lim\left(1+\frac{1}{n-1}\right)^{n-1}\lim\left(1+\frac{1}{n-1}\right)= e$$

Hence,

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

$\endgroup$
1
$\begingroup$

This is a definition of Euler's number as well, I have found a formula that is asymptotic about a week ago while playing around with such limits that is:

$$ \lim_{n \rightarrow \infty} ((n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}) \sim \lim_{n \rightarrow \infty} (1-\frac{1}{n})^{n} $$ where Traian Lalescu has proven that this is true in a rigorous proof which is a more fun way of proofing but takes alot more work

also you can simplify $\frac{n-1}{n}$ = $\frac{n}{n+1}$ which you can use the known limit

$$ \lim_{n \rightarrow \infty} (\frac{n+1}{n})^n = e $$

link:

http://mathhelpforum.com/differential-geometry/102384-solved-prove-lim-1-1-n-n-1-e.html

$\endgroup$
0
$\begingroup$

We have that $$(\frac{n-1}{n})^n=(1-\frac{1}{n})^n=(1+\frac{-1}{n})^n$$ We have the classical limit $$\lim_{x\to\infty}(1+\frac{k}{n})^n=e^k$$ our $k=-1$ so $$\lim_{x\to\infty}(1+\frac{-1}{n})^n=e^{-1}=\frac{1}{e}$$

$\endgroup$
0
$\begingroup$

You can even do

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

With the last equality holding because we are taking the limit as $n$ approaches infinity. Thus, we have

$$\lim\left(\frac{1^{n}}{e}\right) = \frac{1}{e}$$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.