# Arithmetically showing that $$\displaystyle{\frac{{{\log{{\left({x}+{1}\right)}}}}}{{{\log{{\left({x}\right)}}}}}}{ < }{\frac{{{x}+{1}}}{{{x}}}}$$ Is there

Arithmetically showing that $\frac{\mathrm{log}\left(x+1\right)}{\mathrm{log}\left(x\right)}<\frac{x+1}{x}$
Is there a possibility that this can be shown arithmetically? By arithmetically, I mean not looking at the graph.
$\frac{\mathrm{log}\left(x+1\right)}{\mathrm{log}\left(x\right)}<\frac{x+1}{x}$
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alifutlessect692
The derivative of $x↦\frac{\mathrm{ln}x}{x}$ is $\frac{1-\mathrm{ln}x}{{x}^{2}}$ and this is negative if $x>e$ . Thus is $1, we have $\frac{\mathrm{ln}\left(x+1\right)}{x+1}>\frac{\mathrm{ln}x}{x}$ and after dividing by the positive number $\mathrm{ln}x$ and multiplying with $x+1$, this yields $\frac{\mathrm{ln}\left(x+1\right)}{\mathrm{ln}x}>\frac{x+1}{x}$
However, for $x>e$ (and even some smaller x) we obtain $\frac{\mathrm{ln}\left(x+1\right)}{\mathrm{ln}x}<\frac{x+1}{x}$
###### Not exactly what you’re looking for?
Loyau97
Your statement is false in general, as already pointed out. However, the following inequality holds
$\mathrm{log}\left(x+1\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(1+\frac{1}{x}\right)\le \mathrm{log}\left(x\right)+\frac{1}{x}.$
So if $x>e$ then
$\frac{\mathrm{log}\left(x+1\right)}{\mathrm{log}\left(x\right)}\le 1+\frac{1}{x\mathrm{log}\left(x\right)}<1+\frac{1}{x}.$