求和函数,高等数学
2个回答
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x = 0 时, S(x) = 0;
x ≠ 0 时,
S(x) = (1/x) ∑<n=1,∞>(n+1)x^n
= (1/x) ∑<n=1,∞> [x^(n+1)]'
= (1/x) [ ∑<n=1,∞> x^(n+1) ] '
= (1/x) [x^2/(1-x)]'
= (1/x) [x(2-x)/(1-x)^2]
= (2-x)/(1-x)^2 (-1<x<1)
则
S(x) = (2-x)/(1-x)^2, x ∈(-1, 0)∪(0, 1)
S(x) = 0, x = 0
x ≠ 0 时,
S(x) = (1/x) ∑<n=1,∞>(n+1)x^n
= (1/x) ∑<n=1,∞> [x^(n+1)]'
= (1/x) [ ∑<n=1,∞> x^(n+1) ] '
= (1/x) [x^2/(1-x)]'
= (1/x) [x(2-x)/(1-x)^2]
= (2-x)/(1-x)^2 (-1<x<1)
则
S(x) = (2-x)/(1-x)^2, x ∈(-1, 0)∪(0, 1)
S(x) = 0, x = 0
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