关于数列收敛的专题讨论

$f命题：$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

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$f命题：$设连续函数列$left{ {{f_n}left( x ight)} ight}$在$Uleft( {{x_0},delta } ight)left( {delta  > 0} ight)$上一致收敛，且$lim limits_{x o egin{array}{*{20}{c}}{{x_0}} end{array}} {f_n}left( x ight) = {a_n},n in {N_ + }$，证明：数列${a_n}$收敛

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

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$f命题：$设数列${a_n} = sumlimits_{k = 1}^n {frac{k}{{1 + {k^2}}} - ln frac{n}{{sqrt 2 }}} $，证明：数列$left{ {{a_n}} ight}$收敛且极限$ain [0,1/2]$

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

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

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$f命题：$

附录

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

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