Suppose we have a sequence defined by $a_{n+1}=\ln(a_{n}+2)$. We want to prove that for every $a_{0}>0$, the sequence converges to the same $g\in\mathbb{R}$.
This is where I got so far:
Let's define a function $f:[0,+\infty)\to\mathbb{R}$, $f(x)=\ln(x+2)$. First, let's note that $|f^{\prime}(x)|\leq 1/2$ for every $x\geq{0}$. From this, it follows that the inequality $|f(x)-f(y)|\leq\frac{1}{2}|x-y|$ holds for every $x,y\geq{0}$.
Now, let's return to our sequence. One can notice that the terms of the sequence become arbitrarily close to each other:
$\left|a_{2}-a_{1}\right|=\left|f\left(a_{1}\right)-f\left(a_{0}\right)\right|\leq\frac{1}{2}\left|a_{1}-a_{0}\right|$
$\left|a_{3}-a_{2}\right|=\left|f\left(a_{2}\right)-f\left(a_{1}\right)\right|\leq\frac{1}{2}\left|a_{2}-a_{1}\right|=\frac{1}{2}\left|f\left(a_{1}\right)-f\left(a_{0}\right)\right|\leq\left(\frac{1}{2}\right)^{2}\left|a_{1}-a_{0}\right|$
$\vdots\qquad\qquad\vdots$
$\left|a_{n+1}-a_{n}\right|\leq\frac{1}{2}\left|a_{n}-a_{n-1}\right|\leq\left(\frac{1}{2}\right)^{n}\left|a_{1}-a_{0}\right|$
Now, we have:
$\left|a_{n+2}-a_{n}\right|\leq\left|a_{n+2}-a_{n+1}\right|+\left|a_{n+1}-a_{n}\right|$
$\leq\left(\frac{1}{2}\right)^{n}\left|a_{1}-a_{0}\right|+\left(\frac{1}{2}\right)^{n+1}\left|a_{1}-a_{0}\right|$
$=\left[\left(\frac{1}{2}\right)^{n}+\left(\frac{1}{2}\right)^{n+1}\right]\left|a_{1}-a_{0}\right|$
$\vdots\qquad\qquad\vdots$
$\left|a_{m}-a_{n}\right|\leq\left[\left(\frac{1}{2}\right)^{n}+\left(\frac{1}{2}\right)^{n+1}+...+\left(\frac{1}{2}\right)^{m-1}\right]\left|a_{1}-a_{0}\right|$
$=\left(\frac{1}{2}\right)^{n}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+...+\left(\frac{1}{2}\right)^{m-1-n}\right]\left|a_{1}-a_{0}\right|$
$<\left(\frac{1}{2}\right)^{n}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+...\right]\left|a_{1}-a_{0}\right|$
$=\left(\frac{1}{2}\right)^{n}\cdot\frac{1}{1-\frac{1}{2}}\cdot\left|a_{1}-a_{0}\right|$
$=\left(\frac{1}{2}\right)^{n-1}\left|a_{1}-a_{0}\right|$
Thus, for every $\varepsilon>0$ we can find such $N$ that $\left(\frac{1}{2}\right)^{N-1}\left|a_{1}-a_{0}\right|<\varepsilon$. Such $N$ guarantees that for $m,n>N$ the inequality $|a_{m}-a_{n}|<\varepsilon$ also holds. This means that the sequence $a_{n}$ is Cauchy, which also means that it is convergent. This is where my dilemma starts: I believe that this just proves that for every $a_{0}>0$, the sequence $a_{n}$ is convergent to some limit, not necessarily the same one for every $a_{0}$. Am I wrong, and if not, then which path should I take right now?
Thanks in advance.