Xn=(3n+1)/(3n-1)数列的极限Xn=(3n+1)/(3n-1)怎么看她是发散还是收敛啊?并且怎么求该数列的极限?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 05:22:56
![Xn=(3n+1)/(3n-1)数列的极限Xn=(3n+1)/(3n-1)怎么看她是发散还是收敛啊?并且怎么求该数列的极限?](/uploads/image/z/4502142-54-2.jpg?t=Xn%3D%EF%BC%883n%2B1%EF%BC%89%2F%EF%BC%883n-1%EF%BC%89%E6%95%B0%E5%88%97%E7%9A%84%E6%9E%81%E9%99%90Xn%3D%EF%BC%883n%2B1%EF%BC%89%2F%EF%BC%883n-1%EF%BC%89%E6%80%8E%E4%B9%88%E7%9C%8B%E5%A5%B9%E6%98%AF%E5%8F%91%E6%95%A3%E8%BF%98%E6%98%AF%E6%94%B6%E6%95%9B%E5%95%8A%3F%E5%B9%B6%E4%B8%94%E6%80%8E%E4%B9%88%E6%B1%82%E8%AF%A5%E6%95%B0%E5%88%97%E7%9A%84%E6%9E%81%E9%99%90%3F)
Xn=(3n+1)/(3n-1)数列的极限Xn=(3n+1)/(3n-1)怎么看她是发散还是收敛啊?并且怎么求该数列的极限?
Xn=(3n+1)/(3n-1)数列的极限
Xn=(3n+1)/(3n-1)
怎么看她是发散还是收敛啊?并且怎么求该数列的极限?
Xn=(3n+1)/(3n-1)数列的极限Xn=(3n+1)/(3n-1)怎么看她是发散还是收敛啊?并且怎么求该数列的极限?
这个数列是收敛
令{an}为一个数列,且A为一个固定的实数,如果对于任意给出的b>0(b为无限小),存在一个正整数N,使得对于任意n>N,有|an-A|
极限就等于n前的系数之比,
就是看他有没有极限,化解一下可以得,Xn=1+2除以(3n-1),当N无穷大时,2除以(3n-1)为0,即极限为1,即收敛。