神马作文网求下题的极限,利用级数
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/12 04:01:12
求下题的极限,利用级数
收敛必有a>1
令s(x)=∑{1,∞}n*x^(n-1)/a^n
=∑{0,∞}(n+1)*x^n/a^(n+1)
∫{0,x}s(x)dx=∫{0,x}[∑{0,∞}(n+1)*x^n/a^(n+1)]dx
=∑{0,∞}[∫{0,x}(n+1)*x^n/a^(n+1)dx]
=∑{0,∞}(x/a)^(n+1)
=x/a*∑{0,∞}(x/a)^n
=x/a*1/(1-x/a)
=x/(a-x)
s(x)=[x/(a-x)]'=a/(a-x)^2
原极限=s(1)
=a/(a-1)^2
令s(x)=∑{1,∞}n*x^(n-1)/a^n
=∑{0,∞}(n+1)*x^n/a^(n+1)
∫{0,x}s(x)dx=∫{0,x}[∑{0,∞}(n+1)*x^n/a^(n+1)]dx
=∑{0,∞}[∫{0,x}(n+1)*x^n/a^(n+1)dx]
=∑{0,∞}(x/a)^(n+1)
=x/a*∑{0,∞}(x/a)^n
=x/a*1/(1-x/a)
=x/(a-x)
s(x)=[x/(a-x)]'=a/(a-x)^2
原极限=s(1)
=a/(a-1)^2