Showing that $\lim_{n \rightarrow \infty}\left(\frac{n - 1}{n}\right)^n = 1/e$

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.

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.

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 $$\lim_{n\rightarrow\infty}\frac{\frac{n}{n-1}\frac{1}{n^2}}{\frac{-1}{n^2}} = -\lim_{n\rightarrow\infty}\frac{n}{n-1 } = -1$$

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

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,

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

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

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

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 $$\lim_{x\to\infty}(1+\frac{-1}{n})^n=e^{-1}=\frac{1}{e}$$
$$\lim \left(\frac{n-1}{n}\right)^{n} = \lim \left(\frac{1}{\frac{n}{n-1}}\right)^{n} = \lim \left({\frac{1}{\frac{n+1}{n}}}\right)^{n}$$
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.