How can I solve this absolute value logarithmic inequality?

Question

I am trying to solve the following inequality: I would like to show that the absolute value of $1- r \ln (1- \frac 1 r)+ \ln ⁡(1- \frac 1 r)$ is less than $1.$ I already know that $r$ is greater than $0,$ and for this equation to be defined I know that $r$ is greater than $1.$ I can't seem to get anywhere when attempting to prove this. Any help is appreciated. Also apologies for not knowing how to format the equation to look nice on here.

Answer 1

It's wrong. For example, try $r=2$.

Answer 2

If you simplify it you can see that there is no value for $r$ so that the inequality holds: $|1-r\ln(1-1/r)+\ln(1-1/r)| <1=|1+(r-1)\ln(r/r-1)| < 1$, then:

$-1 < 1+(r-1)\ln(r/r-1) <1$, then from right inequality you will obtain:

$(r-1)\ln(r/r-1) < 0$ since r is positive $r-1$ should be positive as well because of the domain of the function $\ln$. So $r>1$. On the other hand, $r/r-1 <1$ so that the inequality $(r-1)\ln(r/r-1)$ becomes negative (or satisfied) which gives 0 < -1 which is a contradiction.

Answer 3

Consider the case where $r$ is large and use Taylor expansion to get $$1-r \log \left(1-\frac{1}{r}\right)+\log \left(1-\frac{1}{r}\right)=2-\frac{1}{2 r}-\frac{1}{6 r^2}+O\left(\frac{1}{r^3}\right)$$

Edit

After Yves Daoust's good comment, using the infinite series $$1-r \log \left(1-\frac{1}{r}\right)+\log \left(1-\frac{1}{r}\right)=2-\sum_{n=1}^\infty\frac{1}{n(n+1)r^n}$$

Answer 4

From the well-known inequality $\ln t\le t-1$, we draw

$$\color{green}{1+(r-1)\ln\frac r{r-1}\le}1+(r-1)\left(\frac r{r-1}-1\right)=\color{green}2.$$

This value is reached at infinity, so the bound is tight.