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
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
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
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}. $
I prefer pictures to words, where possible...
Plot of $x \mapsto (\frac{x+1}{x}-\frac{\log(x+1)}{\log x})$.