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Given that $x_0$ is the unique positive solution of $(2-x)^{n+1}=x(x+1)\cdots(x+n)$, try to find the asymptotic value of $$ M=\prod_{k=0}^n\left(\frac{k+2}{k+x_0}\right)^{k+2} $$ with absolute error $o(1)$ as $n\to\infty$, where $H_n$ denotes $n$-th harmonic number $\sum_{k=1}^n1/k$.

Source (background) IMO2012 problem 2

My answer, which I'm not sure whether is right, is posted isolatedly.

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If you have a complete answer for your own question, you should post it as an answer. –  tomasz Jul 15 '12 at 14:14
    
@tomasz It's okay. –  Frank Science Jul 15 '12 at 14:46
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1 Answer

up vote 2 down vote accepted

All $O$-notations and $o$-notations work for $n\to\infty$.

First we have $2^{n+1}\ge n!x_0$, hence $x_0\le2^{n+1}/n!=O(2^n/n!)$. Take logarithm, we derive that \begin{equation} \begin{split} (n+1)\ln(2-x_0)&=\ln x_0+\sum_{k=1}^n\ln(x_0+k)\\ &=\ln n!+\ln x_0+\sum_{k=1}^n\ln\left(1+\frac{x_0}k\right) \end{split} \end{equation} therefore \begin{gather} \ln(2-x_0)=\ln2+\ln(1-x_0/2)=\ln2+O(x_0)\\ \sum_{k=1}^n\ln(1+x_0/k)=\sum_{k=1}^nO(x_0/k)=O(x_0H_n) \end{gather} \begin{equation} \begin{split} \ln x_0&=(n+1)\ln(2-x_0)-\sum_{k=1}^n\ln\left(1+\frac{x_0}k\right)-\ln n!\\ &=(n+1)(\ln 2+O(x_0))-\ln n!-O(x_0H_n)\\ &=-\ln n!+(n+1)\ln 2+O(nx_0) \end{split} \end{equation} \begin{equation} x_0=\frac{2^{n+1}(1+O(nx_0))}{n!} \end{equation} Notice that $nx_0=O(2^n/(n-1)!)=o(1)$, so we have $x_0\sim2^{n+1}/n!$, thus \begin{equation} x_0=\frac{2^{n+1}}{n!}+O(nx_0^2) \end{equation} Now we can observe $x_0$ more closely \begin{equation} \begin{split} \ln(2-x_0)&=\ln2+\ln(1-x_0/2)=\ln2-x_0/2+O(x_0^2)\\ &=\ln2-2^n/n!+O(nx_0^2) \end{split} \end{equation} \begin{equation} \begin{split} \sum_{k=1}^n\ln(1+x_0/k) &=\sum_{k=1}^n(x_0/k+O(x_0/k)^2)\\ &=H_nx_0+O\left(x_0\sum_{k\ge1}1/k^2\right)\\ &=H_n(2^{n+1}/n!+O(nx_0^2))+O(x_0^2)\\ &=\frac{2^{n+1}H_n}{n!}+O(x_0^2\cdot n\log n) \end{split} \end{equation} \begin{equation} \ln x_0=-\ln n!+(n+1)\ln2-\frac{2^n(n+2H_n+1)}{n!}+O(n^2x_0^2) \end{equation} \begin{equation} \begin{split} x_0&=\frac{2^{n+1}}{n!}\exp\left(-\frac{2^n(n+2H_n+1)}{n!}\right)(1+O(n^2x_0^2))\\ &=\frac{2^{n+1}}{n!}\left(1-\frac{2^n(n+2H_n+1)}{n!}\right)(1+O(n^2x_0^2))\\ &=\frac{2^{n+1}}{n!}\left(1-\frac{2^n(n+2H_n+1)}{n!}\right)+O(n^2x_0^3) \end{split} \end{equation} Next, we compute $\ln M$ \begin{equation} \begin{split} \ln M&=\sum_{k=0}^n(k+2)(\ln(k+2)-\ln(k+x_0))\\ &=\sum_{k=1}^{n+2}k\ln k-\sum_{k=0}^n(k+2)\ln(k+x_0) \end{split} \end{equation} where \begin{equation} \begin{split} \sum_{k=0}^n(k+2)\ln(k+x_0)&=2\ln x_0+\sum_{k=1}^n(k+2)(\ln k+\ln(1+x_0/k))\\ &=2\ln x_0+\sum_{k=1}^n(k+2)\ln k+\sum_{k=1}^n(k+2)\ln(1+x_0/k)\\ &=2\left((n+1)\ln(2-x_0)-\sum_{k=1}^n\ln(1+x_0/k)-\ln n!\right)\\ &\qquad+\sum_{k=1}^n(k+2)\ln k+\sum_{k=1}^n(k+2)\ln(1+x_0/k)\\ &=2(n+1)\ln(2-x_0)+\sum_{k=1}^nk\ln k+\sum_{k=1}^nk\ln(1+x_0/k) \end{split} \end{equation} thus \begin{multline} \ln M=(n+1)\ln(n+1)+(n+2)\ln(n+2)\\ -2(n+1)\ln(2-x_0)-\sum_{k=1}^nk\ln(1+x_0/k) \end{multline} therefore \begin{equation} x_0^2=\frac{4^{n+1}}{n!^2}(1+O(nx_0))^2=\frac{4^{n+1}}{n!^2}+O(nx_0^3) \end{equation} \begin{equation} \begin{split} \ln(2-x_0)&=\ln2+\ln(1-x_0/2)=\ln2-x_0/2-x_0^2/8+O(x_0^3)\\ &=\ln2-\frac{2^n}{n!}\left(1-\frac{2^n(n+2H_n+1)}{n!}\right)-\frac18\frac{4^{n+1}}{n!^2}+O(n^2x_0^3)\\ &=\ln2-\frac{2^n}{n!}+\frac{4^n}{n!^2}\left(n+2H_n+\frac12\right)+O(n^2x_0^3) \end{split} \end{equation} \begin{equation} \begin{split} \sum_{k=1}^nk\ln(1+x_0/k)&=\sum_{k=1}^nk(x_0/k-x_0^2/2k^2+O(x_0/k)^3)\\ &=nx_0-\frac12H_nx_0^2+O\left(x_0^3\sum_{k\ge1}1/k^2\right)\\ &=n\frac{2^{n+1}}{n!}\left(1-\frac{2^n(n+2H_n+1)}{n!}\right)-\frac12H_n\frac{4^{n+1}}{n!^2}+O(n^3x_0^3)\\ &=2n\frac{2^n}{n!}-\frac{4^n}{n!^2}(2n^2+4nH_n+2n+2H_n)+O(n^3x_0^3) \end{split} \end{equation} We have enough stuff to estimate $\ln M$ now. \begin{multline} \ln M=(n+1)\ln(n+1)+(n+2)\ln(n+2)-2(n+1)\ln2\\ +\frac{2^{n+1}}{n!}-\frac{4^n}{n!^2}(n+2H_n+1)+O(n^3x_0^3) \end{multline} thus \begin{equation} \begin{split} M&=\frac{(n+1)^{n+1}(n+2)^{n+2}}{4^{n+1}}\\ &\qquad\qquad\exp\left(\frac{2^{n+1}}{n!}\right)\exp\left(-\frac{4^n}{n!^2}(n+2H_n+1)\right)\\ &\qquad\qquad(1+O(n^3x_0^3)) \end{split} \end{equation} Finally, we have \begin{gather} \exp\left(\frac{2^{n+1}}{n!}\right)=1+\frac{2^{n+1}}{n!}+\frac{2\cdot4^n}{n!^2}+O(x_0^3)\\ \exp\left(-\frac{4^n}{n!^2}(n+2H_n+1)\right)=1-\frac{4^n}{n!^2}(n+2H_n+1)+O(n^2x_0^4) \end{gather} and \begin{equation} \begin{split} M&=\frac{(n+1)^{n+1}(n+2)^{n+2}}{4^{n+1}}\\ &\qquad\qquad\left(1+\frac{2^{n+1}}{n!}-\frac{4^n}{n!^2}(n+2H_n-1)\right)\\ &\qquad\qquad(1+O(n^3x_0^3)) \end{split} \end{equation} Notice that the absolute error \begin{equation} O\left(\frac{n^3(n+1)^{n+1}(n+2)^{n+2}}{4^{n+1}}x_0^3\right) \end{equation} approaches $0$ when $n\to\infty$.

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