求解级数的和函数,划线部分不知怎么得出,望解答下,谢谢
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这里应是有误, 从上页看出,最后一项应是
S3 = ∑<n=0,∞>x^(2n)/(2n+1) = (1/x)∑<n=0,∞>x^(2n+1)/(2n+1)
= (1/x) S4
S4 = ∑<n=0,∞>x^(2n+1)/(2n+1)
(S4)'(x) = ∑<n=0,∞>x^(2n) = 1/(1-x^2), |x| < 1
S4(x) = ∫<0, x>(S4)'(t)dt + S4(0) = ∫<0, x>dx/(1-x^2) + 0
= (1/2)∫<0, x>[1/(1-x)+1/(1+x)]dx = (1/2)ln[(1+x)/(1-x)]
S(3) = [1/(2x)]ln[(1+x)/(1-x)]
? 处应为 (1/x) ∫<0, x>∑<n=0,∞>[t^(2n+1)/(2n+1)]' dt
S3 = ∑<n=0,∞>x^(2n)/(2n+1) = (1/x)∑<n=0,∞>x^(2n+1)/(2n+1)
= (1/x) S4
S4 = ∑<n=0,∞>x^(2n+1)/(2n+1)
(S4)'(x) = ∑<n=0,∞>x^(2n) = 1/(1-x^2), |x| < 1
S4(x) = ∫<0, x>(S4)'(t)dt + S4(0) = ∫<0, x>dx/(1-x^2) + 0
= (1/2)∫<0, x>[1/(1-x)+1/(1+x)]dx = (1/2)ln[(1+x)/(1-x)]
S(3) = [1/(2x)]ln[(1+x)/(1-x)]
? 处应为 (1/x) ∫<0, x>∑<n=0,∞>[t^(2n+1)/(2n+1)]' dt
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