若lim_(ntoinfty)a_(n)=0,则(a_{n)}必为单调递减序列。

A. 对

B. 错

参考答案与解析:

相关试题

2 给出以下4个命题①若lim_(ntoinfty)a_(n)=a,则当n充分大时,|a_(n)-a|0,当n充分大时,|a_(n)-a|0,当n充分大时,|a_(n)-a|

2 给出以下4个命题①若lim_(ntoinfty)a_(n)=a,则当n充分大时,|a_(n)-a|0,当n充分大时,|a_(n)-a|0,当n充分大时,|a

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

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

  • 查看答案
  • 4.设f在[a,b]上有界,(a_{n)}⊂[a,b],lim_(ntoinfty)a_(n)=c.证明:若f在[a,b]上只有a_(n)(n=1,2,...)为其间断点,则f在[a,b]上可积.

    4.设f在[a,b]上有界,(a_{n)}⊂[a,b],lim_(ntoinfty)a_(n)=c.证明:若f在[a,b]上只有a_(n)(n=1,2,...)

  • 查看答案
  • 3 已知数列(a_{n)}(a_(n)≠0),若(a_{n)}发散,则

    3 已知数列(a_{n)}(a_(n)≠0),若(a_{n)}发散,则A. $a_{n}+\frac{1}{a_{n}}$发散.B. $a_{n}-\frac{

  • 查看答案
  • 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)^inftyA_(i),则有P(A)=lim_(

    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

  • 查看答案
  • 2 已知数列 a_{n)} (a_(n) neq 0),若 a_{n)} 发散,则()

    2 已知数列 a_{n)} (a_(n) neq 0),若 a_{n)} 发散,则()A. $\{a_{n} + \frac{1}{a_{n}}\}$ 发散B.

  • 查看答案
  • 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_

    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

  • 查看答案
  • 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_

    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

  • 查看答案
  • 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_(n+1))=sum_(n=1)^inft

    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_(

  • 查看答案
  • 3.[填空题]若lim_(ntoinfty)x_(n)=alpha,则lim_(ntoinfty)|x_(n)|=____.

    3.[填空题]若lim_(ntoinfty)x_(n)=alpha,则lim_(ntoinfty)|x_(n)|=____.3.[填空题]若$\lim_{n\t

  • 查看答案