Let the initial value $x_0 \in (0, 1/2)$.
What is the asymptotic behavior of the converging series $\{x_n\} $?
I have tried the ansatz that
$$x_n \simeq \frac{1}{n} + \frac{a_2}{n^2} + \frac{a_3}{n^3} + \ldots $$
The finding is that $a_2$ can be arbitrary, while there is no consistent value for $a_3$. This indicates that the ansatz is wrong.
The recursion you are interested in can be rewritten as $$\frac1{x_{n+1}}-\frac1{x_n}=\frac1{1-x_n}\tag{$\ast$}$$ You said you already know that $x_n=o(1)$, hence $(\ast)$ yields $$\frac1{x_{n+1}}-\frac1{x_n}=1+o(1)$$ which shows that $$\frac1{x_n}=n+o(n)$$ hence $$x_n=\frac1n+o\left(\frac1n\right)$$ Using this and plugging it back into the RHS of $(\ast)$ yields $$\frac1{x_{n+1}}-\frac1{x_n}=\frac1{1-\frac1n+o\left(\frac1n\right)}=1+\frac1n+o\left(\frac1n\right)$$ which shows that $$\frac1{x_n}=n+\log n+o(\log n)$$ or, equivalently, $$x_n=\frac1n-\frac{\log n}{n^2}+o\left(\frac{\log n}{n^2}\right)$$ This expansion explains why trying to expand $x_n$ into powers of $\frac1n$ fails.
Note finally that one can again plug this last expansion above back into $(\ast)$ above to compute the next terms of the expansion of $\frac1{x_n}$, and consequently of $x_n$.
Let $y_n=1/x_n$. Then $$y_{n+1}=\frac{y_n}{1-1/y_n}=y_n\left(1+\frac1{y_n}+\frac1{y_n^2}+\cdots\right)=y_n+1+O(1/y_n).$$ Therefore $y_n\ge n-A$ for some fixed $A$. Putting this in the above gives $y_n=n+O(\ln n)$. This gives $$x_n=\frac1n+O\left(\frac{\ln n}{n^2}\right).$$ One can go further....
