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f(x) = ∑ x^n/(n+1)xf(x) = ∑ [x^(n+1)]/(n+1)[xf(x)]' = ∑ x^n所以[xf(x)]'的和函數很好求,就是等比級數,所以[xf(x)]' = 1/(1-x)所以xf(x) = ∫ 1/(1-x)dx = -ln(1-x) f(x)=-[ln(1-x)]/x,最後協商收斂於x屬...
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