S(x) = ∑<n=0,∞> x^n/(n+1) = (1/x)∑<n=0,∞> x^(n+1)/(n+1) (x≠0)
S(x) = 0 (x=0)
记 G(x) = ∑<n=0,∞> x^(n+1)/(n+1),
则 G'(x) = ∑<n=0,∞> x^n = 1/(1-x), (-1<x<1)
得 G(x) = ∫<0,x> dt/(1-t) = -ln(1-x), (-1≤x<1)
S(x) = -ln(1-x)/x (-1≤x<1)
S(-1) = ln2
追问G'(x) = ∑ x^n = 1/(1-x), (-1 dt/(1-t) = -ln(1-x), (-1≤x<1)
微分在-1点处无意义,为何积分后在-1点处就有意义了呢,大神能解释解释么
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