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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}$?

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    Sorry to hear it's closed. I$f$ it is reopened, I'll continue.2011-08-15

1 Answers 1

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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.

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    @Srivatsan: The question was closed, so I didn't revisit the answer. However, the asymptotic expansion given yields the requested inequality. I was simply going to add more terms.2012-02-21