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Show that $\log(x+1)-\log(x)<\frac{1}{x}$ for $x >0$.

I was told this was very easy, but I don't see how to solve it. Any help would be great.

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6 Answers 6

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The function $\log(1+t)$ is strictly concave and therefore its graph stays under its tangent line at $0$: for any $t\not=0$ and $t>-1$, $$\log(1+t)< t.$$ Your inequality is equivalent to $$\log(x+1)-\log(x)=\log\left(1+\frac1x\right)< \frac1x.$$

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  • $\begingroup$ @Nicolas FRANCOIS See my edited answer. Anyway I like yours (+1)! $\endgroup$
    – Robert Z
    Dec 21, 2016 at 16:54
  • $\begingroup$ thank you, I must be sleepy :-) And I teach my students to recognize those identities :-P $\endgroup$ Dec 21, 2016 at 16:57
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Apply the mean value theorem: if $f$ is differentiable on $]a,b[$ and continuous on $[a,b]$, then $$(\exists c\in]a,b[)\,f(b)-f(a)=(b-a)f'(c)$$ So here : $$\ln(x+1)-\ln(x)=(x+1-x)\ln'(c)=\frac1c< \frac1x$$ because $x+1>c>x$.

Rmk : you can also study the function $f:x\mapsto \ln(x+1)-\ln(x)-\frac1x$...

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Consider $f(x)=\log\left(\frac{x+1}{x}\right)=\log\left(1+\frac{1}{x}\right)$ . Set $u=1/x$ and then after making a good figure you will see that $\log(1+u)\le u$ and you are done.

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Let $x>0$.

$f: t\mapsto \ln(t)$ is continuous at $[x,x+1]$ and differentiable at $(x,x+1)$, thus by MVT,

$\exists c\in(x,x+1)\;:\; $

$f(x+1)-f(x)=(x+1-x)f'(c)$

$\implies \ln(x+1)-\ln(x)=\frac{1}{c}$

and

$$x<c<x+1 \iff \frac{1}{x+1}<\frac{1}{c}<\frac{1}{x}.$$

$$\implies \frac{1}{x+1}<\ln(x+1)-\ln(x)<\frac{1}{x}$$

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You can see this with $\enspace\displaystyle \frac{1}{x}-\ln(x+1)+\ln x=\int\limits_x^{x+1}(\frac{1}{x}-\frac{1}{t})dt>0\enspace$ because of $\enspace\displaystyle \frac{1}{x}-\frac{1}{t}>0\enspace$ for $\enspace t>x$ .

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If $t\geq 0$,We have:$$ e^t\geq t+1$$ So:$$t\geq \ln{(t+1)}$$ Let:$t=\frac{1}{x}$ then:$$\ln{(1+\frac{1}{x})}\leq \frac{1}{x}$$

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  • $\begingroup$ this is essentially same answer as @Robert Z 's answer, no point in copying answers. $\endgroup$
    – user312097
    Dec 21, 2016 at 17:27
  • $\begingroup$ I'm sorry the answer is repeated, but I did not replicate the answer $\endgroup$
    – Zuo
    Dec 21, 2016 at 17:30
  • $\begingroup$ No problem but in future take care as this can attract downvotes and harsh comments. $\endgroup$
    – user312097
    Dec 21, 2016 at 17:32
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    $\begingroup$ Thank you for your kind reminders, and I think we got this proof in both positive and negative directions $\endgroup$
    – Zuo
    Dec 21, 2016 at 17:34

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