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Is there a possibility that this can be shown arithmetically? By arithmetically, I mean not looking at the graph.

$$\frac{\log(x+1)}{\log(x)} < \frac{x+1}{x}$$

Thank You

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Experiment! Even without a calculator, you can see this must be false if $x$ is a bit bigger than $1$. –  André Nicolas Nov 14 '12 at 19:37

3 Answers 3

up vote 3 down vote accepted

The derivative of $x\mapsto \frac{\ln x}x$ is $\frac{1-\ln x}{x^2}$ and this is negative iff $x>e$. Thus is $1<x< e-1$, we have $\frac{\ln (x+1)}{x+1}>\frac{\ln x}x$ and after dividing by the positive number $\ln x$ and multiplying with $x+1$, this yields $\frac{\ln (x+1)}{\ln x}>\frac{x+1}x$. However, for $x>e$ (and even some smaller $x$) we obtain $\frac{\ln (x+1)}{\ln x}<\frac{x+1}x$.

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Thanks! a perfect answer: true for certain range. Thanks! –  user45099 Nov 14 '12 at 19:44
    
+1 Nice answer. –  copper.hat Nov 14 '12 at 19:45

Your statement is false in general, as already pointed out. However, the following inequality holds

$$\log(x+1) = \log(x) + \log(1+\tfrac{1}{x}) \leq \log(x) + \tfrac{1}{x}. $$

So if $x > e$ then

$$ \frac{\log(x+1)}{\log(x)} \leq 1 + \frac{1}{x \log(x)} < 1 + \frac{1}{x}. $$

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I prefer pictures to words, where possible...

Plot of $x \mapsto (\frac{x+1}{x}-\frac{\log(x+1)}{\log x})$.

enter image description here

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