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2006PTN最后一题的加强改编(西西)

陈洪葛 posted @ 11 年前 in 数学分析 , 1099 阅读

k是一个大于1的整数,且x0>0,满足xn+1=xn+1kxn,求
lim
求证
\lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}
证明:
我们将要证明
  \lim_{n\to\infty}\frac{x^{k+1}_{n}}{n^{k}}=\left(\frac{k+1}{k}\right)^{k}
为此,只要证明
  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}
而归纳得到
x_{n+1}>x_{n},x_{n}>0
因此,x_{n}的单调递增序列。设A=\lim_{n\to+\infty}x_{n},则有A=+\infty.说明x_{n}\to+\infty,n\to+\infty.
(x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}})=x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}} \right)^{1+\frac{1}{k}}-1\right)=x_{n}^{1+\frac{1}{k}}\left(\exp\left[\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}} \right) \right]-1 \right)\sim \frac{k+1}{k}(n\to+\infty)
所以,由O.Stolz定理
  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}
对于加强版本,我们先看一个事实,对任意一个收敛的序列,比如说
\lim_{n\to\infty}a_{n}=A
那么,
\lim_{n\to\infty}\frac{(a_{n}^{k}-A^{k})}{a_{n}-A}=kA^{k-1}
这样,设a_{n}=\frac{x_{n}^{\frac{k+1}{k}}}{n},就有\displaystyle \lim_{n\to\infty}a_{n}=\frac{k+1}{k}=A.
\lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}^{k}-A^{k})= \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)\cdot k\cdot \left(\frac{k+1}{k}\right)^{k-1}
于是,只要证
\lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)=\frac{1}{2k}
就是
\lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{1}{2k}
\frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}
这时,可以用O.Stolz定理了。就有
\begin{align*} &\lim_{n\to\infty}\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\\ &\overbrace{=}^{O.Stolz}\lim_{n\to\infty}\frac{x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}}-A}{\ln\left(1+\frac{1}{n} \right)}\\ &=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}}\right)^{1+\frac{1}{k}}-1\right) -A\right]\\ &=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}}\right) } -1\right)-A\right]\\ &=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}+o\left( \left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right) \right) }-1 \right) -A\right]\\ &=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{k+1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)+\frac{1}{2}\left(\frac{k+1}{k}\right)^{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)^{2} \right)-A\right]\\ &=\lim_{n\to\infty}\frac{k+1}{2k^{2}}\frac{n}{x_{n}^{1+\frac{1}{k}}}\\ &=\frac{1}{2k} \end{align*}
因此,命题得证。


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