# Show $\lim\limits_{n\to\infty} n \ln\left(1-\frac{1}n\right) = -1$

Could you help me show that $$\lim\limits_{n\to\infty} n \ln \left({1-\frac{1}n} \right) = -1 ?$$

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## 6 Answers

Recall that $$\lim_{n \to \infty}\left( 1 - \dfrac1n \right)^n = e^{-1}$$ Since $\log$ is a continuous function, we have that $$\lim_{n \to \infty} n \log\left( 1 - \dfrac1n \right) = \lim_{n \to \infty} \log\left( 1 - \dfrac1n \right)^n = \log \left( \lim_{n \to \infty}\left( 1 - \dfrac1n \right)^n \right) = \log \left(e^{-1} \right) = -1$$

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I really like this. – Potato Dec 29 '12 at 21:53
+1 nice proof as usual from Marvis! – doniyor Dec 29 '12 at 21:54
Nice! Would it make sense to add some more brackets around the log (on the first and second step) to make it more readable? – leo Dec 29 '12 at 21:59

Recall that for $x>0$, $$1-\frac{1}{x} \leq \log x \leq x-1.$$ Then $$\frac{-n}{n-1} \leq n \log \left(1-\frac{1}{n}\right) \leq -1,$$ and the result follows by taking $n \rightarrow \infty$.

The first inequality can be derived from $\log x= \int_1^x \frac{dt}{t}$, since $$\int_1^x \frac{dt}{t^2} \leq \int_1^x \frac{dt}{t} \leq \int_1^x dt.$$

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Another way, and the way I would do it in practice:

We have the power series expansion

$\log(1-1/x)=-\frac{1}{x}+O\left(\frac{1}{x}^2\right)\dots$

So

$x\log(1-1/x)=-1+O\left(\frac{1}{x}\right)$

and letting $x$ tend to infinity gives the answer.

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A Potato has shown me the power of O-notation . . . Never thought that'd happen. +1 – 000 Dec 29 '12 at 21:59
@Limitless well, the space of possible events is limitless. (not the same as infinite, I know...) – kram1032 Dec 29 '12 at 23:13

$$\lim\limits_{n\to\infty} n \ln\left({1-\frac{1}n}\right)=\lim_{n\to\infty} \ln\left({1-\frac{1}n}\right)^n=\ln\lim_{n\to\infty}\left(1-{1\over n}\right)^n=\ln e^{-1}=-1$$

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Using L'Hopital rule, $$\lim\limits_{n\to\infty} n \ln \left({1-\frac{1}n} \right) = \lim_{n \rightarrow \infty} \frac{ \ln \left(1 - \frac 1 n \right)}{ \frac 1 n} = \lim_{n \rightarrow \infty} \frac{\frac{1}{1 - \frac 1 n}}{- \frac 1 {n^2}} \times \frac{1}{n^2} = -1$$

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I think the $-1 / n^2$ just before the last equal sign should not have a minus sign. – pimvdb Dec 29 '12 at 22:17
@pimvdb yes thank you for correction!! – Santosh Linkha Dec 29 '12 at 22:20

We take the limit:

$\lim\limits_{n\to\infty} n \ln\left({1-\frac{1}n}\right)$

This is a limit of the type '0 $\infty$'. Let $t=\frac{1}n$:

$=\lim\limits_{t\to 0} \dfrac{\ln\left({1-t}\right)}t$

This is a limit of the type '0/0'. We apply L'Hospital's rule:

$=\lim\limits_{t\to 0} \dfrac{1}{t-1}$

The limit of the quotient is defined if it is defined for numerator and denominator. The limit of the constant '1' in the numerator is the constant '1'.

$=\dfrac{1}{\lim\limits_{t\to 0} t-1}$

The answer is therefore

$= -1$

(Solved with the help of wolfram alpha)

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