Show that $\log(x+1)-\log(x)<\frac{1}{x}$ for $x >0$ 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.
 A: 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
$$\log(x+1)-\log(x)=\log\left(1+\frac1x\right)< \frac1x.$$
A: 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 $f:x\mapsto \ln(x+1)-\ln(x)-\frac1x$...
A: 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.
A: 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
$$x<c<x+1 \iff   \frac{1}{x+1}<\frac{1}{c}<\frac{1}{x}.$$
$$\implies \frac{1}{x+1}<\ln(x+1)-\ln(x)<\frac{1}{x}$$
A: 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 $\enspace t>x$ .
A: If $t\geq 0$,We have:$$ e^t\geq t+1$$ So:$$t\geq \ln{(t+1)}$$  Let:$t=\frac{1}{x}$  then:$$\ln{(1+\frac{1}{x})}\leq \frac{1}{x}$$
