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
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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