Let's say we have the equation $$a_{n+1} = Frac\left(\frac{1}{\ln(1+a_n)}\right)$$ with $a_0\ne 0$ and that it is not negative. I've tried finding the limit of $a_n$ as $n \rightarrow \infty$, but I wasn't able to find an answer. I would like to ask what the answer would be, and how you would find it.
$Frac(x)$ is the fractional part.
On
If $y_n = 1/\log(1+a_n)$, you have $y_{n+1} = m + a_{n+1}$ for some integer $m$. There will be solutions for every positive integer $m$, where $y_n \to y$ with $y = 1/\log(y + 1 - m)$ and $m \le y \le m+1$. For example, for $m=1$ we have $y \approx 1.763222834$ and for $m=2$ we have $y \approx 2.493404089$. These correspond to $a_n \to 0.763222834$ and $a_n \to 0.493404089$ approximately.
BTW, for $m=1$ the solution to $y = 1/\log(y)$ is $1/W(1)$ where $W$ is the Lambert W function. The other cases don't have "closed form" solutions AFAIK.
Suppose that $(a_n)$ has a limit, say $a$. Then, by the continuity of $t\mapsto 1/t$ and $t\mapsto \log(1+t)$ we have that $a=1/\log(a+1)$. This equation has exactly two solutions, one in the interval $(1,2)$ and one in the interval $(-1,0)$. (you can see that with elementary calculus, e.g. intermediate value theorem). Since $a_n\geq 0$, the limit cannot be in the interval $(-1,0)$ and therefore is the solution that lies in $(1,2)$.