Consider $a_n = \ln(1 + a_{n-1})$ with $a_0 = 1/2$. Show the limit is $0$ and find the radius of convergence (proof verification) Consider $a_n = \ln(1 + a_{n-1})$ with $a_0 = \frac{1}{2}$. Show the limit is $0$ and find the radius of convergence
I have seen these:
Find limit of $a_n = \ln (1 + a_{n-1})$
Convergence rate of $a_{n+1}=\ln(a_n+1)$
But on both the initial condition is different, and the second one seems to be asking for something different.
First, let's find the limit. I argued that $a_n$ is monotone decreasing.
In general, $e^x \geq 1 + x$ unless $x = 0$. But, $a_0 = 1/2$, so $a_n \neq 0 \ \forall x$. Therefore, I can say:
$e^{a_n} \geq 1 + a_n$
$a_n \geq \ln(1 + a_n) = a_{n+1}$
Since $a_n \geq 0$ and $a_n$ is decreasing, we can conclude that its limit is $0$.
Does this look ok? I have not worked very much with recursive sequences, so I want to make sure I'm doing ok.
Now, let's work on the radius of convergence.
$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}$ is what we want to compute. I am going to drop the limit as I won't actually end up needing it for what I attempted (which is a bit of a red flag for me, but it's all I could think to try).
$\frac{a_{n+1}}{a_n} = \frac{\ln(1+a_n)}{\ln(1+a_{n-1})} = \ln(a_n - a_{n-1}) = \ln(\ln(a_{n-1} - a_{n-2})) = \dots = \ln(\ln(\dots(\ln(a_1 - a_0)))\dots) = \ln(\ln(\dots(\ln(a_1 - \frac{1}{2})))\dots)$
Now, I already showed that $a_n$ was decreasing. So, $a_1 - \frac{1}{2} <0$ and this means the limit does not exist. Now, I know the $\limsup_{n \to -\infty} \ln(n) = \infty$. Making a change of variable, $\limsup_{n \to \infty} \ln(-n) = \infty$ still. Therefore, I believe the limsup is $\infty$ which means $R = 0$.
Does everything look ok here? Did I mess up on any parts?
 A: 
In general, $e^x \geq 1+x$ unless $x = 0$

I guess you meant to say that $e^x > 1+x$? Now, you actually should explicitly show that $a_n > 0$ for all $n \in \mathbb{N}_0$. The proof is just a simple induction argument.
Indeed, $a_0 > 0$. Next, assume that $a_n > 0$ for some arbitrary $n \in \mathbb{N}_0$. Then:
$$a_{n+1} = \log(a_{n+1}+1) > 0$$
since $\log$ is a strictly increasing function. Your conclusion that your sequence is decreasing is correct and since it is bounded below, it is convergent with limit $L$. It would, then, follow that:
$$L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \log(1+a_n) = \log(1+L)$$
due to the continuity of $L$. This is only possible when $L = 0$. That would be a much more complete argument.
As for your computation of the radius of convergence, you've overcomplicated it significantly. It does not follow that:
$$\frac{\log(1+a_n)}{\log(1+a_{n-1})} = \log(a_n-a_{n-1})$$
Since every $a_n > 0$, we have that:
$$\frac{1}{R} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\log(1+a_n)}{a_n} = \lim_{n \to \infty} \frac{\log(1+a_n)-\log(1)}{a_n}$$
By the sequential limit criterion, the right-hand side is just the derivative of $x \mapsto \log(x)$ at $x = 1$. So, it follows that $R = 1$.
