关于数列收敛的专题讨论

$\bf命题：$设${x_{n + 1}} = \cos {x_n},n = 0,1,2, \cdots $

   (1)利用上下极限证明${x_n}$收敛到$x = \cos x$的解

   (2)利用${x_2n}$和${x_(2n+1)}$的单调有界性证明${x_n}$收敛到$x=\cos x$的解

   (3)利用${x_n}$为Cauchy列来证明${x_n}$收敛到$x=\cos x$的解

1

$\bf命题：$设连续函数列$\left\{ {{f_n}\left( x \right)} \right\}$在$U\left( {{x_0},\delta } \right)\left( {\delta  > 0} \right)$上一致收敛，且$\lim \limits_{x \to \begin{array}{*{20}{c}}{{x_0}} \end{array}} {f_n}\left( x \right) = {a_n},n \in {N_ + }$，证明：数列${a_n}$收敛

1

$\bf命题：$设$f(x)$是$\left[ {1, + \infty } \right)$上的非负单调减少函数，令${a_n} = \sum\limits_{k = 1}^n {f\left( k \right)} - \int_1^n {f\left( x \right)dx} ,n \in {N_ + }$，证明：数列$\left\{ {{a_n}} \right\}$收敛

1

$\bf命题：$设数列${a_n} = \sum\limits_{k = 1}^n {\frac{k}{{1 + {k^2}}} - \ln \frac{n}{{\sqrt 2 }}} $，证明：数列$\left\{ {{a_n}} \right\}$收敛且极限$a\in [0,1/2]$

1

$\bf命题：$设${f_n}\left( x \right) = {e^{\frac{x}{{n + 1}}}},n \in {N_ + }$，数列${y_n}$满足：${y_1} = c > 0,\frac{n}{{n + 1}}\int_0^{{y_{n + 1}}} {{f_n}\left( x \right)dx}  = {y_n}$，求极限${y_n}$

1

$\bf命题：$设$f(x)$可微，且$0 < f'\left( x \right) \leqslant \frac{k}{{1 + {x^2}}}\left( {k > 0} \right)$，对$\forall {x_0} \in R$，令${x_{n + 1}} = f\left( {{x_n}} \right)$，证明：$\left\{ {{x_n}} \right\}$收敛于$f(x)$的不动点

1

$\bf命题：$

附录

$\bf命题：$设$\left\{ {{a_n}} \right\},\left\{ {{b_n}} \right\}$均为正整数数列，且适合$${a_1} = {b_1} = 1,{a_n} + \sqrt 3 {b_n} = {\left( {{a_{n - 1}} + \sqrt 3 {b_{n - 1}}} \right)^2}$$证明：数列$\left\{ {\frac{{{a_n}}}{{{b_n}}}} \right\}$的极限存在，并求其极限值

1

$\bf命题：$设${a_1} = 2,{a_n} = \frac{{1 + \frac{1}{n}}}{2}{a_{n - 1}} + \frac{1}{n}$，证明：$\lim \limits_{n \to \infty } n{a_n}$存在