Find asymptotic of recurrence sequence Given a sequence $x_1=\frac{1}{2}$, $x_{n+1}=x_n-x_n^2$. It's easy to see that it limits to $0$.
The question is: is there exists an $\alpha$ such, that $\lim\limits_{n\to\infty}n^\alpha x_n\neq0$.
I tried to find explicit formula for $x_n$, but did not succeed. I stucked at that point: if $a_1=(\frac{1}{4})^2$, $a_n=(a_{n-1}+\frac{1}{4})^2$, then $x_n=-a_{n-2}+\frac{1}{4}$ for $n\ge3$. 
Thanks for any ideas and help.
 A: Let $y_n = \frac{1}{x_n} - n$, we have
$$y_{n+1} = \frac{1}{x_{n+1}}-(n+1) = \frac{1}{x_n} + \frac{1}{1-x_n} - (n+1) = y_n + \frac{x_n}{1-x_n}$$
This implies for $n > 1$,
$$y_n = y_1 + \sum_{k=1}^{n-1} \frac{x_k}{1-x_k}
\quad\iff\quad
\frac{1}{x_n} = n + 1 + \sum_{k=1}^{n-1}\frac{x_k}{1-x_k}
\tag{*1}$$
Since $0 \le x_k < 1$, this immediately give us an upper bound of $x_n$:
$$\frac{1}{x_n} \ge n+1 \quad\iff\quad x_n \le \frac{1}{n+1}\tag{*2}$$
Substitute $(*2)$ in $(*1)$ and notice the map $x \mapsto \frac{x}{1-x}$ is an increasing function for $x \in [0,1)$, we obtain a lower bound for $x_n$:
$$\begin{align}
 & \frac{1}{x_n} \le n+1 + \sum_{k=1}^{n-1}\frac{\frac{1}{k+1}}{1 - \frac{1}{k+1}}
= n + 1 + \sum_{k=1}^{n-1}\frac{1}{k} \le n + 1 + \log n + \gamma\\
\implies & x_n \ge \frac{1}{n + 1 + \log n + \gamma}\tag{*3}
\end{align}$$
where $\gamma$ is the 
Euler–Mascheroni constant.
Combine these two bounds, it is clear we can take $\alpha = 1$ and $\displaystyle\lim_{n\to\infty} n^\alpha x_n = 1$.
