$ b_{n + 1} = \frac {b_n^2 + 2b_n}{b_n^2 + 2b_n+2}$ and $ b_1 = 1$, show that $ \left|\frac{2}{n}-\frac{2\ln{n}}{n^2}-b_n\right|\leq\frac{1}{n^2}$ In the recursion $ b_{n + 1} = \frac {b_n^2 + 2b_n}{b_n^2 + 2b_n+2}$, with $ b_1 = 1,$ how can one prove that  $ \left|\frac{2}{n}-\frac{2\ln{n}}{n^2}-b_n\right|\leq\frac{1}{n^2}$?
 A: Note that
$$
b_{n+1}+1=2\frac{(b_n+1)^2}{(b_n+1)^2+1}
$$
Letting $b_n=c_n-1$, we get
$$
c_{n+1}=2\frac{c_n^2}{c_n^2+1}
$$
Letting $c_n=\frac{1}{d_n}$, we get
$$
d_{n+1}=\frac{1}{2}(1+d_n^2)
$$
Note that $d_1=\frac{1}{2}$, and if $0\le d_n\le 1$, then $0\le d_{n+1}\le 1$. Thus, $0\le d_n\le 1$ for all $n$.
Letting $d_n=1-e_n$, we get
$$
e_{n+1}=e_n-\frac{1}{2}e_n^2
$$
Note that $e_n$ is non-increasing and $0\le e_n\le 1$ for all $n$. Therefore, $\lim_{n\to\infty}e_n$ exists. Thus, $\lim_{n\to\infty}\;\frac{1}{2}e_n^2=\lim_{n\to\infty}(e_n-e_{n+1})=\lim_{n\to\infty}\;e_n-\lim_{n\to\infty}\;e_{n+1}=0$. Therefore, $\lim_{n\to\infty}\;e_n=0$.
Letting $e_n=\frac{1}{f_n}$ (whereby $\lim_{n\to\infty}f_n=\infty$), we get
$$
\begin{align}
f_{n+1}-f_n&=\frac{1}{2}\frac{f_{n+1}}{f_n}\\
&=\frac{1}{2}\left(1+\frac{f_{n+1}-f_n}{f_n}\right)
\end{align}
$$
Collecting the $f_{n+1}-f_n$ on the left, we get
$$
(f_{n+1}-f_n)\left(1-\frac{1}{2f_n}\right)=\frac{1}{2}
$$
So that
$$
f_{n+1}-f_n=\frac{1}{2}\left(1+\frac{1}{2f_n}+\frac{1}{4f_n^2}+\frac{1}{8f_n^3}+\dots\right)
$$
We can iteratively apply the Euler-Maclaurin Sum Formula starting with $f_n=\frac{a+n}{2}$. Two passses gives
$$
f_n=\frac{1}{2}((a+n)+\log(a+n)+\frac{\log(a+n)}{a+n})+O\left(\frac{1}{a+n}\right)
$$
Note that $b_n=\frac{1}{f_n-1}$. This yields $b_n=\frac{2}{n}-\frac{2\log(a+n)}{n^2}-\frac{2(a-2)}{n^2}+\dots$.
Had a few minutes, so I added a bit more. Gotta go again; I will finish this later.
