Convergence rate of the sequence $a_{n+1} = a_n-a_n^2, a_0=1/2$. The sequence converges to zero at a rate that seems to be slightly faster than $1/n$.
What are the best known results on the convergence rate of this sequence?

Here is a slightly better bound:
first, it is more convenient to look at $b_n = 1/a_n$. We have $$\frac{1}{b_{n+1}}=\frac{1}{{b_n}}-\frac{1}{{b_n^2}}=\frac{b_n-1}{b_n^2}$$ or, $$b_{n+1}=\frac{b_n^2}{b_n-1}=\frac{(b_n+1)(b_n-1)+1}{b_n-1}=b_n+1+\frac{1}{b_n-1}.$$
It is easy to see that $b_k\geq 2$ for all $k$, which implies
$$b_{n}\leq b_{n-1}+1+\frac{1}{2-1}\leq\dots\leq 2n+2.$$
Now,
$$
 b_{n+1}\geq b_n+1+\frac{1}{1+2n}\geq b_{n-1}+2+\frac{1}{2n+1}+\frac{1}{2n-1}\geq \dots\geq b_0+n+1+\sum_{k=0}^n \frac{1}{2k+1},
$$
and since
$$
 \sum_{k=0}^n \frac{1}{2k+1} > \frac{1}{2}\sum_{k=0}^n \frac{1}{k+1} \geq  \frac{1}{2}\ln(n+2)
$$
we get
$$
 b_{n+1}\geq n+\frac{1}{2}\ln(n+2)+3.
$$
Using the same reasoning, from $b_n\geq n+2$ (and using $\sum_{k=1}^n \frac{1}{k} \leq \ln n+1$) we get
$$
 b_{n+1}\leq n+\ln(n+1)+4,
$$
hence
$$
 \frac{1}{n+\ln(n+1)+4}\leq a_{n+1} \leq \frac{1}{n+\frac{1}{2}\ln(n+2)+3}.
$$
