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

Question

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

Answer 1

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

Answer 2

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

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

Answer 3

I prefer pictures to words, where possible...

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