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}$?
$ 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}$
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6Have you tried? Try induction and use the recursion. – 2011-08-14
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5I edited the tags, because I don't think this has much to do with algebra. But: this has the air of another past contest problem. Do you even try to work on these on your own? Show your work, and then there is much better chance of making progress. Sincerely, I don't think that posting many contest questions here in quick succession is the best way to train your abilities. Honestly: if you are training for a contest, I am trying to be helpful here. If this is not a contest problem from the past, then I apologize. – 2011-08-14
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1The function $f(x)=(x^2+2x)/(x^2+2x+2)$ gives the single iteration here. If instead of $f$ the function $$g(x)=\frac{x^2+2x}{(x^2/2)+2x+2}=\frac{2x}{x+2}$$ were used, then starting with $b_1=1$ we would have $b_n=2/(n+1)$ for all $n$. The difference $g(x)-f(x)$ increases about cubically in $x$ from $0$ to $1/15$ in the interval $x\in[0,1]$, so the question is about some kind of stability of the recursively generated sequence under small variations in the function being iterated. I don't know if this helps :-( – 2011-08-14
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3Why does this question has 4 close-votes for "too localized"? The technique used to show this can surely be of interest to others. – 2011-08-14
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1@Jonas: I consider this to be a very nice question, and not trivial, since no fast answer was posted. Sure, if the OP would have explained more about the problem and what he tried, he wouldn't have so many close votes. – 2011-08-14
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1@Jonas: Nothing wrong about this particular question really. The negative feelings are probably a result of OP posting several questions in a short span, and apparently not having done any work on any of them. At least nothing s/he wanted to share with us. Probably won't get closed now that many people have thought about the problem. The votes to close are mostly a message to the OP. I am one of those voters. – 2011-08-14
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1I voted to close with similar reasons as Jyrki and Theo, but would happily vote to reopen if Amir edits in his motivation and what he's tried in solving this. I'll note that similar interventions helped a former offender change his habits... – 2011-08-14
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6@Jonas: It was an (evidently failed) attempt to educate Amir, see my comments [here](http://math.stackexchange.com/questions/57057/challenging-inequality-abcde-1-show-that-frac1a-frac1b-frac1c) and [here](http://math.stackexchange.com/questions/57377/true-or-false-the-largest-k-such-that-akabkb-geq-akbbka-holds). – 2011-08-14
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0@J. M. *similar interventions helped a former offender change his habits*... **REALLY??** This would be terrific news but I missed the case. Until now, I only saw zillions of examples of the opposite. (Surprise, I voted to close. Not really happy about the "too localized" motive though, but one must choose a reason from a limited list.) – 2011-08-14
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1@Didier: It took quite a bit of prodding+downvotes, TBH. I don't want to name names, but a search of meta might net you the name of the former offender. SFAICT he no longer dumps questions. – 2011-08-14
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0Sorry to hear it's closed. If it is reopened, I'll continue. – 2011-08-15
1 Answers
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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0I ran into this thread just now. Just reminding you about the answer, in case you want to finish it. :) – 2012-02-21
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0@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