# How can I solve this absolute value logarithmic inequality?

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.

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

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 $$1-r \log \left(1-\frac{1}{r}\right)+\log \left(1-\frac{1}{r}\right)=2-\sum_{n=1}^\infty\frac{1}{n(n+1)r^n}$$

• Hi Claude. This is a nice, compact approach. Anyway, a Big-O term in the expression doesn't prevent it to exceed $2$ for some finite $r$. – Yves Daoust Feb 27 at 7:53
• @YvesDaoust. Good point ! I added the infinite summation. Moreover, we can study the function the first and second derivatives being always positive. – Claude Leibovici Feb 27 at 8:26

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.

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.