第1个回答 2020-06-10
如图
第2个回答 2020-06-10
可以用root test (faster), 得
lim{n->oo} |x| < 1
-1 < x < 1 (at both boundaries it diverges.)
Since ∑x^n = 1/(1-x) - 1......(1), n from 1 to oo,
Differentiate both sides of (1), ∑nx^n = 1/(1-x)^2......(2), n from 1 to oo
Integrate both sides of ∑x^(n-1) = 1/(1-x) and multiply by 2, ∑(2/n)x^n = -2ln(1-x)......(3)
n from 1 to oo
和函数 = 1/(1-x)^2 - 2ln(1-x)本回答被网友采纳
lim{n->oo} |x| < 1
-1 < x < 1 (at both boundaries it diverges.)
Since ∑x^n = 1/(1-x) - 1......(1), n from 1 to oo,
Differentiate both sides of (1), ∑nx^n = 1/(1-x)^2......(2), n from 1 to oo
Integrate both sides of ∑x^(n-1) = 1/(1-x) and multiply by 2, ∑(2/n)x^n = -2ln(1-x)......(3)
n from 1 to oo
和函数 = 1/(1-x)^2 - 2ln(1-x)本回答被网友采纳
第3个回答 2021-01-02
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