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