Boundedness of the sequence $R_{n+1}=\ln\left(1+a R_n\right),a>1$ I want to show that the sequence $R_{n+1}=\ln\left(1+\frac{\pi}{\lambda} R_n\right),\pi>\lambda>0$ converges, where $R_1=1-\frac{\lambda}{\pi}$. I already showed by induction that it is monotonely increasing but I am struggling with boundedness.
My idea is to do this inductively as well.
However, Calculalting $R_2$ gives me $R_2=\ln\left(\pi/\lambda\right)$. I believe that $\pi$ is a threshhold, but by induction I would get $R_{n+1}=\ln\left(1+\frac{\pi}{\lambda} R_n\right) \le R_n \frac{\pi}{\lambda}$.
Also, I already know that there exists some $R>0$ such that $R=\ln\left(1+R\frac{\pi}{\lambda}\right)$.
 A: Proposition 1. $\{R_n\}$ is ascending.
$f(x)=\ln(1+ax)$ is ascending function for $\forall x \geq 0$, since $f'(x)=\frac{a}{1+ax}>0, \forall x \geq 0$. Now, using this inequality and because $a>1>0$
$$R_2=\ln\left(1+a\left(1-\frac{1}{a}\right)\right)=\ln{a}\geq \frac{a-1}{a}=R_1$$
Given $f(x)$ is ascending, $f(R_2)\geq f(R_1)$ which is $R_3\geq R_2$ and, by induction, $\color{red}{R_{n+1}\geq R_n}$.

Proposition 2. $\ln\left(1+x^2\right) \leq x, \forall x\geq 0$.
$g(x)=x-\ln\left(1+x^2\right)$ is ascending, since $g'(x)=1-\frac{2x}{1+x^2}\geq 0 \iff 1+x^2 \geq 2x \iff (1-x)^2 \geq 0$. Considering $g(0)=0$, then $x \geq 0 \Rightarrow g(x) \geq g(0)=0 \Rightarrow \color{red}{x \geq \ln\left(1+x^2\right)}$.

Proposition 3. $R_n \leq a$.
Using induction:


*

*$R_1=1-\frac{1}{a}<1<a$

*$R_2=\ln{a} < a$

*Let's assume $R_n \leq a$, then $1+aR_n \leq 1+a^2$. Function $\ln{x}$ is ascending, then $\color{red}{R_{n+1}}=\ln\left(1+aR_n\right)\leq \ln\left(1+a^2\right)\color{red}{\leq a}$ from Proposition 2.



Altogether, $\{R_n\}$ is ascending and bounded, thus, it's converging.
A: $r_1 
= 1-\frac1{a}$,
$r_2 
= \ln(1+ a(1-\frac1{a}))
= \ln(a)
$,
$r_3
=\ln(1+a\ln(a))
\lt a \ln(a)
$.
This might be a start.
If
$r_{n+1}
=\ln(1+ar_n)
$
with
$a > 1$
then
$\begin{array}\\
r_{n+2}-r_{n+1}
&=\ln(1+ar_{n+1})-\ln(1+ar_n)\\
&=\ln(\frac{1+ar_{n+1}}{1+ar_n})\\
&=\ln(\frac{1+ar_n-ar_n+ar_{n+1}}{1+ar_n})\\
&=\ln(1+\frac{a(r_{n+1}-r_n)}{1+ar_n})\\
&\lt\frac{a(r_{n+1}-r_n)}{1+ar_n}
\qquad\text{since } \ln(1+x) < x\\
\text{so}\\
\dfrac{r_{n+2}-r_{n+1}}{r_{n+1}-r_n}
&\lt \dfrac{a}{1+ar_n}\\
\end{array}
$
If we can show that
$\dfrac{a}{1+ar_n}
\lt 1$,
we will be done.
However,
it is almost midnight here,
so this is enough for me.
