9.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx,b_(n)=int_(0)^(pi)/(2)sin^ntdt,则极限lim_(ntoinfty)[((n+1)a_(n))/(b_(n))]^n=( ).

9.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx,b_(n)=int_(0)^(pi)/(2)sin^ntdt,则极限lim_(ntoinf

20.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx(n=0,1,2,...),则lim_(ntoinfty)((a_(n))/(a_(n-2

2.[判断题] 判断：int_(0)^1x^m(1-x)^ndx=int_(0)^1x^n(1-x)^mdx,(m,nin N).（）A. 对B. 错

4.证明：int_(0)^1x^m(1-x)^ndx=int_(0)^1x^n(1-x)^mdx(m,nin N).4.证明：$\int_{0}^{1}x^{m

设 M=int_(-(pi)/(2))^(pi)/(2) (sin x)/(1+x^2) cos^4 x dx，N=int_(-(pi)/(2))^(pi)/(

设数列|x_(n)|满足：x_(1)in(0,pi)，x_(n+1)=sin x_(n)(nin N_(+)).证明lim_(ntoinfty)x_(n)存在，

设A_(1),A_(2),...,A_(n),...是事件列，若A_(n)subset A_(n+1),n=1,2,...,A=bigcap_(i=1)^inf

2、设级数sum_(n=1)^inftya_(n)收敛，lim_(ntoinfty)na_(n)=a.证明：sum_(n=1)^inftyn(a_(n)-a_(

(6)设 _(n)=dfrac (3)(2)(int )_(0)^dfrac (n{n+1)}(x)^n-1sqrt (1+{x)^n}dx, 则极限limna

设x_(0)=0,x_(n)=(1+2x_(n-1))/(1+x_(n-1))(n=1,2,3,...),则lim_(ntoinfty)x_(n)=设$x_{0