3. 因 S（0）= 0 则 S(x) = ∑ x^(2n+1)/(2n+1) = ∑ ∫<0, x> t^(2n) dt + S(0) = ∫<0, x> ∑ t^(2n) dt = ∫<0, x> dt/(1-t^2) = (1/2)ln[(1+x)/(1-x)] (-1 < x < 1) ∑ 1/[(2n+1)2^n] = ∑ (1/√2)^(2n) / (2n+1) = √2 ∑ (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。
