An inequality involving recurrence sequence $b_{n+1}=\frac{b_{n}^{2}+2b_{n}}{b_{n}^{2}+2b_{n}+2}$ with $b_{1}=1$ and still unsolved 
Given a recursion $b_{n+ 1}= \dfrac{b_{n}^{2}+ 2b_{n}}{b_{n}^{2}+ 2b_{n}+ 2}$ with $b_{1}= 1.$ Prove that
$$\left | \frac{2}{n}- \frac{2\ln n}{n^{2}}- b_{n} \right |\leq\frac{1}{n^{2}}$$
Source: AoPS/@Ji_Chen (still unsolved)

I think this problem related to the hyperbol functions, how can we find $b_{n}$ or such a relationship between $b_{n}$ and $n.$ Professor Ji always inspires me by beautiful asymptotics, at most, I only used his result to show $\lim b_{n}= 0$ not to use Laurent's series, hope you find more observations involving the recurrence sequence above, I love it.
 A: This is not meant as a complete proof of the inequality above (I've got no motivation to solve a problem without the complete source or information about the importance of exactly that recursion), just a few thoughts how that might be accomplished:
With $r_n=b^{-1}_n$, the recursion becomes (after some elementary algebra)
$$r_{n+1}-r_n=\frac12+\frac1{4\,r_n+2}\tag{1}$$ with $r_1=1$. Now $r_1>0$ implies that all $r_n>0$, so we have $$r_{n+1}-r_n>\frac12,$$ and this, together with $r_1=1$, gives
$$r_n\ge\frac{n+1}2.\tag{2}$$
Plugging (2) into (1), we obtain
$$r_{n+1}-r_n\le\frac12+\frac1{2(n+2)},$$ and by summing up,
$$r_n\le\frac{n+H_{n+1}}2-\frac14,\tag{3}$$ where $$H_n=\sum^n_{k=1}\frac1k$$ is the $n$-th harmonic number. Since $H_n-\ln n\to\gamma$ as $n\to\infty$, the RHS of (3) already gives the first two terms $$\frac2n-\frac{\ln n}{n^2}$$ of the asymptotics of $b_n=r^{-1}_n$, as a lower bound. For the upper bound, and the inequality above, one would have to plug (3) into (1), again, and do some rather technical estimates.
BTW, for $b_1\in(-2,0)$, $b_n$ converges to $-1$.
