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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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    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.2012-06-28

6 Answers 6

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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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    See Edit. $ $ $ $2012-06-28
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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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    Got it! Thanks Andre!2012-06-28
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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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    We have a on going debate for homework. Regardless, this didn't help the OP much, so some extra detail was necessary.2012-06-28
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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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    OK. Good to know2012-06-28
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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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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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    Yeah, it seems he does. But from the mistakes he did, I assumed he was starting with limits of this kind.2012-06-28