Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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 $\iff t=0$.

share|improve this answer

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

share|improve this answer
    
$\geq {1 \over n}$. –  Felix Marin Sep 7 '13 at 1:22
    
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 $n>1$. –  user71352 Sep 7 '13 at 1:26

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 $$f^\prime(x) = 1-\frac{1}{1+x} = \frac{x}{1+x}\geq 0.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.