# Asymptotic behavior of the expression: $(1-\frac{\ln n}{n})^n$ when $n\rightarrow\infty$

The well known results states that:

$\lim_{n\rightarrow \infty}(1-\frac{c}{n})^n=(1/e)^c$ for any constant $c$.

I need the following limit: $\lim_{n\rightarrow \infty}(1-\frac{\ln n}{n})^n$.

Can I prove it in the following way? Let $x=\frac{n}{\ln n}$, then we get: $\lim_{n\rightarrow \infty}(1-\frac{\ln}{n})^n=\lim_{x\rightarrow \infty}(1-\frac{1}{x})^{x\ln n}=(1/e)^{\ln n}=\frac{1}{n}$.

So, $\lim_{n\rightarrow \infty}(1-\frac{\ln}{n})^n=\frac{1}{n}$.

I see that this is wrong to have an expression with $n$ after the limit. But how to show that the asymptotic behavior is $1/n$?

Thanks!

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No, it is not. Finally, a limit should be a number (or infinity), how can you conclude that the limit is $1/n$? –  Siminore Jun 28 '12 at 14:39
You can't have $n$ outside of the limit. Perhaps you wish to argue the function in question is asymptotically $1/n$, but that takes more work; you can't hold one part of the expression fixed, take the limit, and then take the limit of the resulting expression. For instance, you could get $$1=\lim_{x\to0}\frac{x}{x}=\lim_{x\to0}\frac{\lim_{x\to0}x}{x}=\lim_{x\to0} \frac{0}{x}=\lim_{x\to0}0=0.$$ –  anon Jun 28 '12 at 14:45
Thanks! Yes, what I need is to show that the expression $(1-\frac{\ln n}{n})^n$ is asymptotically $1/n$. How I can show this? –  Michael Jun 28 '12 at 14:50
If you change the nature of your question in comments, please change the question, or, if you don't have the permission, ask somebody else to change it. –  Thomas Andrews Jun 28 '12 at 15:44

According to the comments, your real aim is to prove that $x_n=n\left(1-\frac{\log n}n\right)^n$ has a non degenerate limit.

Note that $\log x_n=\log n+n\log\left(1-\frac{\log n}n\right)$ and that $\log(1-u)=-u+O(u^2)$ when $u\to0$ hence $n\log\left(1-\frac{\log n}n\right)=-\log n+O\left(\frac{(\log n)^2}n\right)$ and $\log x_n=O\left(\frac{(\log n)^2}n\right)$.

In particular, $\log x_n\to0$, hence $x_n\to1$, that is, $$\left(1-\frac{\log n}n\right)^n\sim\frac1n.$$

Edit: In the case at hand, one knows that $\log(1-u)\leqslant-u$ for every $u$ in $[0,1)$. Hence $\log x_n\leqslant0$ and, for every $n\geqslant1$, $$\left(1-\frac{\log n}n\right)^n\leqslant\frac1n.$$

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Thank you for the answer! As I asked Andre, what is the correct writing of such an asymptotic behavior? What I actually need is to show that the probability $\Pr(Event)=(1-\frac{\ln n}{n})^n$ goes to $0$ at least as fast as $1/n$. –  Michael Jun 28 '12 at 15:51
See Edit.   –  Did Jun 28 '12 at 16:11
Thanks a lot Did!! –  Michael Jun 28 '12 at 21:21

The idea you used could have worked. Exactly as you wrote, we have $$\left(1-\frac{\ln n}{n}\right)^n=\left(\left(1-\frac{1}{x}\right)^x\right)^{\ln n},$$ where $x=\frac{n}{\ln n}$.

Note that $x\to \infty$ as $n\to\infty$. Note also that $\ln n=g(x)$ for some function $g(x)$ such that $g(x)\to\infty$ as $x\to \infty$. Our limit is $$\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{g(x)}.$$ Since $\left(1-\frac{1}{x}\right)^x$ approaches $\frac{1}{e}$, and $g(x)\to\infty$, our limit is $0$.

Remark: Your impossible answer $\frac{1}{n}$ contained a large kernel of truth. When $n$ is large, $(1-1/x)^x$ is close to $1/e$, so the original expression is close to $1/n$. And of course the quantity $1/n$ approaches $0$.

Let $E(n)$ be the original expression. We can adapt your argument to show that $\lim_{n\to\infty} nE(n)=1$, which proves in a very informative way that $E(n)$ has limit $0$, by giving quite exact information about the rate of approach to $0$.

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Thanks! You mean in the last sentence that $E(n)$ has limit $1/n$? In general, how to correctly denote this asymptotic behavior. Since it is indeed incorrect to write that the limit when $n\rightarrow \infty$ is a function of $n$. –  Michael Jun 28 '12 at 15:48
@Michael: I wrote it down in the comment above, $\lim_{n\to\infty} nE(n)=1$, or if you prefer, equivalently $\lim{n\to\infty} \frac{E(n)}{1/n}=1$. Our expression approaches $0$ like $1/n$. The most natural proof to me involves the Taylor expansion. Let $t=(\log n)/n$. Then $\log(1-t)=-t+O(t^2)$. And I had a typo at the end that I fixed, and prompted your question. the limit of $E(n)$ is $0$, of course. –  André Nicolas Jun 28 '12 at 15:58
Got it! Thanks Andre! –  Michael Jun 28 '12 at 16:01

Hint: consider $$\log \left( \left( 1-\frac{\log n}{n} \right)^n \right) = n \log \left( 1-\frac{\log n}{n} \right)$$ and prove (or recall) that $$\lim_{n \to +\infty} \frac{\log n}{n} =0.$$ Since $\log (1+\varepsilon ) \approx \varepsilon$ as $\varepsilon \to 0$, ...

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No: you get $$n \frac{\log \left( 1- \frac{\log n}{n} \right)}{\frac{\log n}{n}} \frac{\log n}{n} \to -\infty$$ –  Siminore Jun 28 '12 at 14:56
"Since $\log (1+\varepsilon) \approx \varepsilon$ as $\varepsilon \to 0$" –  Siminore Jun 28 '12 at 15:03
Fair enough. =) (I guess you can make \approx a \sim) –  Pedro Tamaroff Jun 28 '12 at 15:15
The choice between $\approx$ and $\sim$ seems to depend on your culture. In Italy, we tend to use $\approx$, but also different symbols that I can't even write in TeX! –  Siminore Jun 28 '12 at 16:42
OK. Good to know –  Pedro Tamaroff Jun 28 '12 at 16:51

Your approach is wrong (in particular, how could your result contain $n$?).

Notice that $(1-\frac{\ln n}{n})^n=e^{n\ln(1-\frac{\ln n}{n})}$ and $\frac{\ln n}{n}\rightarrow 0$, so $n\ln(1-\frac{\ln n}{n})\sim-\ln n\rightarrow -\infty$.

Moreover $\displaystyle\lim_{n\to-\infty}e^n=0$.

Conclusion : $\displaystyle\lim_{n\to+\infty}(1-\frac{\ln n}{n})^n=0$.

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How does it follow that $n \log \left(1-\frac{\log n} n \right) \sim -\log n$ given that $\log n / n \to 0$? I don't think the OP knows about asymptotical behaviour, and it seems you're also using the series expansion of $\log(1-x)$ near $x=0$. –  Pedro Tamaroff Jun 28 '12 at 14:56
Viewing OP's comment, I assume he knows about asymptotical behavior. –  JBC Jun 28 '12 at 15:04
Yeah, it seems he does. But from the mistakes he did, I assumed he was starting with limits of this kind. –  Pedro Tamaroff Jun 28 '12 at 15:12

No, it is incorrect. The answer is 0.

Since $\lim_{n\to\infty}(1-\frac{\ln n}{n})^{\frac{n}{\ln n}}=\frac{1}{e}<1$ and $\lim_{n\to\infty}\ln n=+\infty$, we have that $\lim_{n\to\infty}(1-\frac{\ln n}{n})^n=\lim_{n\to\infty}((1-\frac{\ln n}{n})^{\frac{n}{\ln n}})^{\ln n}=0$.

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Since you're new here, some tips: When answering it is good you be detailed, so: Why is it incorrect? Why is the answer $0$? –  Pedro Tamaroff Jun 28 '12 at 14:42
On the other hand, for homework-type problems like this, maybe it is better to provide just a hint and wait for the OP to try again. –  GEdgar Jun 28 '12 at 14:49
I agree with GEdgar. –  Siminore Jun 28 '12 at 15:05
We have a on going debate for homework. Regardless, this didn't help the OP much, so some extra detail was necessary. –  Pedro Tamaroff Jun 28 '12 at 15:07

Let $A=\lim_{n\to\infty}(1-\ln n/n)^n$. Take $\ln$ both sides,we get $lnA=\lim_{n\to\infty}n ln(1-\ln n/n)$.Now put $x=1/n$,it will give you $lnA=lim_{y\to0^+}\frac{ln(1+y\ln y)}{y}$.Using L'Hopital's Rule,you will get this limit=$-\infty$.Therefore,$\ln A=-\infty \implies A=0$.So, the required limit is $0$.

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