How can you show that $$(1-\frac{1}{n})^r \leq e^{-r/n}\ ?$$
I'm not really too sure how to do it.
Exponential function inequality
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$\begingroup$
real-analysis
inequality
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0can you tell us something about the variables? – 2017-02-19
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0How do you mean? – 2017-02-19
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0what kind of numbers are $$r,n$$? – 2017-02-19
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0Tell us whether $n$ is a natural number or not and so on. – 2017-02-19
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0from where does this inequality come?. – 2017-02-19
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0Oh of course. Both $r$ and $n$ are natural numbers ($\neq 0$). – 2017-02-19
2 Answers
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I showed in THIS ANSWER that the sequence $e_n$, given by $e_n=\left(1-\frac1n\right)^n$ is monotonically increasing. Recalling that $\lim_{n\to \infty}e_n=e^{-1}$ we have
$$\left(1-\frac1n\right)^n whence raising both sides of $(1)$ to the $r/n$, where $r\in \mathbb{N}$, power yields the coveted inequality $$\left(1-\frac1n\right)^r as was to be shown!
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taking the logarithm on both sides we obtain $$r\ln\left(1-\frac{1}{n}\right)<-\frac{r}{n}$$ if $$r>0$$ then we can divide by $$r$$ and we get $$\ln\left(1-\frac{1}{n}\right)<-\frac{1}{n}$$ multipliying by $-1$ we get $$-\ln\left(1-\frac{1}{n}\right)^n>1$$ can you finish this?