Eufisky - The lost book

## 利用导数不等式进行数列求和估计

（1）证明$\{x_n\}$为递减数列;

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

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

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

（2） 先证明左边不等式.首先有

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

$\sqrt {{x_n}} - \frac{1}{{\sqrt {{x_n}} }} > \ln {x_n}.$

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

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

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

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

## 解题过程中碰到的几个特殊数列

1.Catalan数

$a_n=a_0a_{n-1}+a_1a_{n-2}+\cdots+a_{n-1}a_0,\quad n\geq1,$

\begin{align*}F\left( x \right) = &xf\left( x \right) = {a_0}x + {a_1}{x^2} +  \cdots + {a_n}{x^{n + 1}} +  \cdots \\{F^2}\left( x \right) = &a_0^2{x^2} + \left( {{a_0}{a_1} + {a_1}{a_0}} \right){x^3} +  \cdots \\&+ \left( {{a_0}{a_{n - 1}} + {a_1}{a_{n - 2}} +  \cdots  + {a_{n - 1}}{a_0}} \right){x^{n + 1}} +  \cdots \\= &{a_1}{x^2} + {a_2}{x^3} +  \cdots  + {a_n}{x^{n + 1}} +  \cdots \\= & F\left( x \right) - {a_0}x,\end{align*}

$\sqrt {1 - 4x} = 1 + C_{1/2}^1\left( { - 4x} \right) + C_{1/2}^2{\left( { - 4x} \right)^2} + \cdots + C_{1/2}^{n + 1}{\left( { - 4x} \right)^{n + 1}} + \cdots ,$

\begin{align*}{a_n} &=  - \frac{1}{2}C_{1/2}^{n + 1}{\left( { - 4} \right)^{n + 1}}\\&=  - \frac{1}{2}\frac{{\frac{1}{2} \cdot \left( { - \frac{1}{2}} \right) \cdot \left( { - \frac{3}{2}} \right) \cdots \left( { - \frac{{2n - 1}}{2}} \right)}}{{n!}}{\left( { - 1} \right)^{n + 1}}{4^{n + 1}} = \frac{{C_{2n}^n}}{{n + 1}}.\end{align*}

2.Bell数

\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty  {\frac{{{n^k}}}{{n!}}}  = \sum\limits_{n = 1}^\infty  {\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}}  = \sum\limits_{n = 0}^\infty  {\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\&= {b_{k - 1}} + C_{k - 1}^1{b_{k - 2}} +  \cdots  + C_{k - 1}^{k - 2}{b_1} + {b_0},\end{align*}

$B\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{B\left( n \right)}}{{n!}}{x^n}} = {e^{{e^x} - 1}}.$

$$\mathrm e^{\mathrm e^x} = \sum_{k=0}^\infty\frac{\mathrm e^{kx}}{k!} = \sum_{k=0}^\infty\frac 1{k!}\biggl(\sum_{n=0}^\infty\frac{(kx)^n}{n!}\biggr) = \sum_{n=0}^\infty\frac{x^n}{n!}\biggl(\sum_{k=0}^\infty\frac{k^n}{k!}\biggr),$$

$$a_n=\frac 1{n!}\sum_{k=0}^\infty\frac{k^n}{k!} > \frac{k^n}{n!\,k!},$$

$$\frac {k^n}{n!\,k!}=\frac{\displaystyle\Bigl(\frac n{\ln n}\Bigr)^n}{n!\,\displaystyle\Bigl(\frac n{\ln n}\Bigr)!} \sim\frac{\displaystyle\Bigl(\frac n{\ln n}\Bigr)^n}{\displaystyle\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n\cdot \sqrt{\frac{2\pi n}{\ln n}}\Bigl(\frac n{\mathrm e\ln n}\Bigr)^{\frac n{\ln n}}} =\frac{(\mathrm e\ln n)^{\frac n{\ln n}}\sqrt{\ln n}}{2\pi n(\ln n)^n}.$$