如题所述
f(x)
=
∑
x^n/(n+1)
xf(x)
=
∑
[x^(n+1)]/(n+1)
[xf(x)]'
=
∑
x^n
所以[xf(x)]'的和函数很好求,就是等比级数,所以
[xf(x)]'
=
1/(1-x)
所以xf(x)
=
∫
1/(1-x)dx
=
-ln(1-x)
f(x)=-[ln(1-x)]/x,
最后协商收敛于x属于[-1,0)
U
(0,1)采纳哟
=
∑
x^n/(n+1)
xf(x)
=
∑
[x^(n+1)]/(n+1)
[xf(x)]'
=
∑
x^n
所以[xf(x)]'的和函数很好求,就是等比级数,所以
[xf(x)]'
=
1/(1-x)
所以xf(x)
=
∫
1/(1-x)dx
=
-ln(1-x)
f(x)=-[ln(1-x)]/x,
最后协商收敛于x属于[-1,0)
U
(0,1)采纳哟
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