记
f(x) = ∑(n=2~∞)[nx^(n-1)]/(n-1)
= ∑(n=2~∞)x^(n-1) + ∑(n=2~∞)[x^(n-1)]/(n-1)
= g(x) + h(x),
利用已知级数
∑(n=1~∞)x^(n-1) = 1/(1-x) ,-1<x<1,
可得
g(x) = ∑(n=2~∞)x^(n-1) = ∑(n=1~∞)x^(n-1) - 1 = 1/(1-x) - 1,-1<x<1,
h‘(x) = {∑(n=2~∞)[x^(n-1)]/(n-1)}’
= {∑(n=2~∞)x^(n-2)
= 1/(1-x) ,-1<x<1,
积分,得
h(x) = ∫[0,x]h'(t)dt = ∫[0,x][1/(1-t)]dt
= -ln(1-x),-1<x<1,
于是
f(x) = ……。
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