高等数学，2 3求和函数
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3. 因 S（0）= 0
则 S(x) = ∑<n=0,∞> x^(2n+1)/(2n+1) = ∑<n=0,∞> ∫<0, x> t^(2n) dt + S(0)
= ∫<0, x> ∑<n=0,∞> t^(2n) dt = ∫<0, x> dt/(1-t^2) = (1/2)ln[(1+x)/(1-x)]
(-1 < x < 1)
∑<n=0,∞> 1/[(2n+1)2^n] = ∑<n=0,∞> (1/√2)^(2n) / (2n+1)
= √2 ∑<n=0,∞> (1/√2)^(2n+1) / (2n+1)
= (√2/2)ln[(1+1/√2)/(1-1/√2)] = (√2/2)ln[(√2+1)/(/√2-1)]
= (√2/2)ln[(√2+1)^2] = √2 ln(√2+1)
上题更简单，自行仿做即可。
则 S(x) = ∑<n=0,∞> x^(2n+1)/(2n+1) = ∑<n=0,∞> ∫<0, x> t^(2n) dt + S(0)
= ∫<0, x> ∑<n=0,∞> t^(2n) dt = ∫<0, x> dt/(1-t^2) = (1/2)ln[(1+x)/(1-x)]
(-1 < x < 1)
∑<n=0,∞> 1/[(2n+1)2^n] = ∑<n=0,∞> (1/√2)^(2n) / (2n+1)
= √2 ∑<n=0,∞> (1/√2)^(2n+1) / (2n+1)
= (√2/2)ln[(1+1/√2)/(1-1/√2)] = (√2/2)ln[(√2+1)/(/√2-1)]
= (√2/2)ln[(√2+1)^2] = √2 ln(√2+1)
上题更简单，自行仿做即可。
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