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How do I show that

$$ \frac{1}{n-1}\geq \ln \left ( \frac{n}{n-1} \right ) $$

for $ n>1 $?

As far as I can tell, exponentiating both sides with base $e$ won't help, because then I get a nasty term on the LHS.

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We know that for any $t\geqslant 0$ $$\log(1+t)\leqslant t$$ since when $t\geqslant 0$ $$\int_0^t \frac 1{1+x}dx\leqslant\int_0^t dx$$ Then take $t=x^{-1}$. Note equality is true $\iff t=0$.

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For $n>1$ we have $\ln(\frac{n}{n-1})=\ln(n)-\ln(n-1)=\int_{n-1}^{n}\frac{1}{x}dx\le\frac{1}{n-1}$.

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  • $\begingroup$ $\geq {1 \over n}$. $\endgroup$ – Felix Marin Sep 7 '13 at 1:22
  • $\begingroup$ I'm sorry did I make a mistake? I used that $n-1\le x\le n$ so $\frac{1}{n}\le \frac{1}{x}\le\frac{1}{n-1}$ for $n>1$. $\endgroup$ – user71352 Sep 7 '13 at 1:26
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Note that $\frac{n}{n-1} = \left(1+\frac{1}{n-1}\right)$, so in essence we need to prove $x\geq\ln(1+x)$ for all $x\geq 0$.

Well, if we let $f(x) = x-\ln(1+x)$ then $f(0)=0$ and for all $x>0$ we have $$f^\prime(x) = 1-\frac{1}{1+x} = \frac{x}{1+x}\geq 0.$$

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Let's notice, that $$\Big( 1+\frac{1}{n} \Big)^{n}<e<\Big(1+\frac{1}{n} \Big)^{n+1}, n \in \text{1, 2, 3, ... , n, ...}$$ $$\ln\Big(1+\frac{1}{n} \Big)^{n}<1<\ln\Big(1+\frac{1}{n} \Big)^{n+1}$$ $$\ln\Big(1+\frac{1}{n} \Big)^{n}<1 \text{ that gives us } \ln\Big(1+\frac{1}{n} \Big)<\frac{1}{n}$$ and $$\ln\Big(1+\frac{1}{n} \Big)^{n+1}>1 \text{ that gives us } \ln\Big(1+\frac{1}{n} \Big)>\frac{1}{n+1}$$ Hope this will help.

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