利用导数不等式进行数列求和估计 - Eufisky - The lost book
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利用导数不等式进行数列求和估计

Eufisky posted @ 2015年12月25日 03:20 in 数学分析 with tags 数列 , 1354 阅读

设$x_1=\frac{p+1}{p},p>1,x_{n+1}=\frac{x_n-1}{\ln x_n}$.

(1)证明$\{x_n\}$为递减数列;

(2)证明\[\frac{1}{{p + 1}} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}} < \ln \left( {{x_1}{x_2} \cdots {x_n}} \right) < \frac{1}{p} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}}.\]


证.(1)首先利用$\ln(1+x)\leq (x>-1)$可知

\[{x_{n + 1}} = \frac{{{x_n} - 1}}{{\ln {x_n}}} \ge \frac{{{x_n} - 1}}{{{x_n} - 1}} = 1.\]

又$x_1=\frac{p+1}{p}>1$,显然有$x_n>1$.再利用$\ln (1+x)\geq \frac x{x+1}$,我们有

\[\frac{{{x_{n + 1}}}}{{{x_n}}} = \frac{{{x_n} - 1}}{{{x_n}\ln {x_n}}} < \frac{{{x_n} - 1}}{{{x_n} \cdot \frac{{{x_n} - 1}}{{{x_n}}}}} = 1.\]

因此$\{x_n\}$为递减数列.


(2) 先证明左边不等式.首先有

\[\frac{{\ln {x_{n + 1}}}}{{\ln {x_n}}} = \frac{{\ln {x_{n + 1}}}}{{\ln {x_n}}} = \frac{{\ln \left( {\frac{{{x_n} - 1}}{{\ln {x_n}}}} \right)}}{{\ln {x_n}}} > \frac{1}{2},\]

\[\sqrt {{x_n}} - \frac{1}{{\sqrt {{x_n}} }} > \ln {x_n}.\]

这只要在不等式$\frac{1}{2}\left( {x - \frac{1}{x}} \right) > \ln x,x > 1$中令$x=\sqrt{x_n}>1$便可得到.又利用$\frac x{x+1}\leq \ln (x+1)\leq x$可知\[\frac{1}{{p + 1}} < \ln {x_1} = \ln \left( {1 + \frac{1}{p}} \right) < \frac{1}{p}.\]

于是

\[\ln {x_n} > \frac{1}{2}\ln {x_{n - 1}} > \cdots > \frac{1}{{{2^{n - 1}}}}\ln {x_1} > \frac{1}{{p + 1}} \cdot \frac{1}{{{2^{n - 1}}}}.\]

将上述不等式前$n$项累加便可得到

\[\frac{1}{{p + 1}} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}} < \ln \left( {{x_1}{x_2} \cdots {x_n}} \right).\]


接着证明右边不等式.由不等式

\[\ln x > \frac{{2\left( {x - 1} \right)}}{{x + 1}},\quad x > 1\]

可知\[{x_{n + 1}} = \frac{{{x_n} - 1}}{{\ln {x_n}}} < \frac{{{x_n} + 1}}{2}.\]

因此\[{x_{n + 1}} - 1 < \frac{{{x_n} - 1}}{2} \Rightarrow {x_n} \le 1 + \frac{1}{{p{2^{n - 1}}}} \Rightarrow \ln {x_n} < \frac{1}{{p{2^{n - 1}}}},\]

从而有

\[\ln \left( {{x_1}{x_2} \cdots {x_n}} \right) < \frac{1}{p} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}}.\]

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