So the sequence is monotonically decreasing. So it clearly suffices to find $B$ such that $a_n \le \frac{B}{n}$ holds whenever $n=2^k$. By induction, we have \[a_{m+d} \le a_m-A(a_m^2+a_{m+1}^2+\dots+a_{m+d-1}^2) \le a_m-dAa_{m+d}^2,\]i.e. \[a_{m+d}(1+dAa_{m+d}) \le a_m.\]In particular, putting $m=d$ we have \[a_{2m}(1+mAa_{2m}) \le a_m.\]By rescaling the sequence appropriately, we may assume that $A=2$. Writing $b_n=2^na_{2^n}$ we conclude that \[b_{n+1}(1+b_{n+1}) \le 2b_n.\]Hence $b_{n+1} \le b_n$ for $b_{n+1} \ge 1$ and $b_{n+1}^2 \le b_n$ for $b_{n+1} \le 1$. We need to prove that the sequence $b_n$ is bounded. Indeed, assume that $b_n \le 1$ for some $n$, then $b_k \le 1$ for all $k \ge n$ and we are done. So we may suppose that $b_n > 1$ for all $n$. But then $b_{n+1} \le b_n$ for all $n$ and the sequence is decreasing, hence bounded.

Tintarn wrote: So the sequence is monotonically decreasing. So it clearly suffices to find $B$ such that $a_n \le \frac{B}{n}$ holds whenever $n=2^k$. By induction, we have \[a_{m+d} \le a_m-A(a_m^2+a_{m+1}^2+\dots+a_{m+d-1}^2) \le a_m-dAa_{m+d}^2,\]i.e. \[a_{m+d}(1+dAa_{m+d}) \le a_m.\]In particular, putting $m=d$ we have \[a_{2m}(1+mAa_{2m}) \le a_m.\]By rescaling the sequence appropriately, we may assume that $A=2$. Writing $b_n=2^na_{2^n}$ we conclude that \[b_{n+1}(1+b_{n+1}) \le 2b_n.\]Hence $b_{n+1} \le b_n$ for $b_{n+1} \ge 1$ and $b_{n+1}^2 \le b_n$ for $b_{n+1} \le 1$. We need to prove that the sequence $b_n$ is bounded. Indeed, assume that $b_n \le 1$ for some $n$, then $b_k \le 1$ for all $k \ge n$ and we are done. So we may suppose that $b_n > 1$ for all $n$. But then $b_{n+1} \le b_n$ for all $n$ and the sequence is decreasing, hence bounded. Can u plz explain these 2 things - (1.) Why it suffices to prove it for $n=2^k$ (2.) And how can u assume $A=2$ ?