Cynthia Bell

2021-12-31

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

Shawn Kim

The function $\mathrm{log}\left(1+t\right)$ is strictly concave and therefore its graph stays under its tangent line at $0:$ for any $t\ne 0$ and $t>-1$,
$\mathrm{log}\left(1+t\right)
$\mathrm{log}\left(x+1\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(1+\frac{1}{x}\right)<\frac{1}{x}$

Edward Patten

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

Vasquez

Let $x>0$.
$f:t\to \mathrm{ln}\left(t\right)$ is continuous at $\left[x,x+1\right]$ and differentiable at $\left(x,x+1\right)$, thus by MVT,
$\mathrm{\exists }c\in \left(x,x+1\right):$
$f\left(x+1\right)-f\left(x\right)=\left(x+1-x\right)f\prime \left(c\right)$
$⇒\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x\right)=\frac{1}{c}$
and
$x
$⇒\frac{1}{x+1}<\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x\right)<\frac{1}{x}$

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