How to find monotony of $x_{n+1} = \ln(1 + x_n)$

Question

How do I find the monotony of the series $x_{n}=\ln(1+x_{n-1}),x_{1}=a, 0<a<1$?

I know I should do the difference between $x_{n+1}-x_{n}$ and see if it is positive or negative but I get $\ln\frac{1+x_{n}}{1+x_{n-1}}$ and I do not know what to do after. Can you help me?

Answer 1

$$x_{n+1}-x_n=\ln(1+x_n)-x_n<0$$

Let $f(x)=x-\ln{(1+x)}$, where $x>-1$.

Thus, $f'(x)=1-\frac{1}{x+1}=\frac{x}{x+1}$, which says that $x_{\min}=0$.

That is, $f(x)\geq f(0)=0$.

In our case $x>0$. Thus, $\ln(1+x)-x<0$.