利用导数不等式进行数列求和估计 - Eufisky - The lost book
FoxTrot Series
利用定积分证明组合恒等式

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 , 1308 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

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

Avatar_small
mi tv samsung no se 说:
2023年7月30日 04:57

Es posible que el Wi-Fi no funcione en su televisor Samsung pero esté funcionando en otros dispositivos, o puede estar conectado al punto de acceso de su teléfono y acceder a Internet de esa manera, mi tv samsung no se conecta a internet pero se niega a conectarse a Internet a través de su enrutador. Esta guía simple lo ayudará a volver a conectar su televisor Samsung en línea si no se conecta a Internet pero otros dispositivos pueden hacerlo o si Wi-Fi no funciona en absoluto.

Avatar_small
KVS 6th Class Sylla 说:
2023年9月01日 23:12

Kendriya Vidyalaya (KVS) 6th Class Students Regular Reading new Syllabus for Half Yearly, Final Exam Easy to Pass Curriculum for the Academic Year 2024, KVS 6th class Students Download your KVS 6th Split Up Syllabus 2024 Online Pdf Format.This year KVS has Announced 6th Class Split Up Syllabus for for the subjects English, Hindi, Mathematics, Sanskrit, Science, SST, German etc. have KVS 6th Class Syllabus 2024 a look at the Exam Pattern and Syllabus for KVS 8th Class Curriculum for the Academic Year 2024,Kendriya Vidyalaya (KVS) 6th Class Curriculum Syllabus & Exam Pattern Download Regular Redding Best Performs in for Home Work & Final Exam Curriculum for the Academic Year 2024.


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter