设S(x)=∑[x^(2n)]/(2n+1),S(x)=∑[x^(2n+1)]'=[∑x^(2n+1)]'=[(x^3)/(1-x^2)]'=(3x^2-x^4)/(1-x^2)^2∑1/(2n+1)2^n=∑[(1/√2)^2n]/(2n+1)=S(1/√2)=(3*(1/2)-(1/4))/(1-(1/2))^2=5答案是√2ln(√2+1)再算算吧呵呵设S(x)=∑[x^(2n)]/(2n+1)|x|<1xS(x)=∑[x^(2n+1)]/(2n+1)[xS(x)]'=∑x^(2n)=x²/(1-x²)xS(x)=∫x²/(1-x²)dx=∫[1/(1-x²)-1]dx=(1/2)∫[1/(1-x)+1/(1+x)]dx-x+C=(1/2)ln|(1+x)/(1-x)|-x+Cx=0时，xS(x)=0*S(0)=(1/2)ln|(1+0)/(1-0)|-0+C C=0∑1/(2n+1)2^n=∑[(1/√2)^2n]/(2n+1)=S(1/√2)x=1/√2时,xS(x)=[(1/2)ln|(1+1/√2)/(1-1/√2)|]-1/√2(1/√2)S(1/√2)=ln(1+√2)-1/√2S(1/√2)=√2ln(1+√2)-1为什么我算出的结果多一个常数---------------------------------------------------换种方法：xS(x)=∑[x^(2n+1)]/(2n+1)=(x^3)/3+(x^5)/5+(x^7)/7+……ln(1+x)=x-(x^2)/2+(x^3)/3-(x^4)/4+……-ln(1-x)=x+(x^2)/2+(x^3)/3+(x^4)/4+……可见xS(x)=(1/2)[ln(1+x)-ln(1-x)]-x得到的是和上面方法一样的结果，楼主的题设恐怕有误，如果n是从0到∞而不是从1到∞，那么才能计算出楼主给的那个结果对 确实是写错了呵呵不好意思，应该是0到无穷。那么以上几步应该改为什么呢？法1中第3行[xS(x)]'=∑x^(2n)=x²/(1-x²)应改为∑x^(2n)=1/(1-x²)，应为首项为1，而不是x²，法2中xS(x)=∑[x^(2n+1)]/(2n+1)=x+(x^3)/3+(x^5)/5+(x^7)/7+……那么xS(x)=(1/2)[ln(1+x)-ln(1-x)](1/√2)S(1/√2)=[(1/2)ln|(1+1/√2)/(1-1/√2)|]=ln(1+√2)S(1/√2)=√2ln(1+√2)∑1/(2n+1)2^n=∑[(1/√2)^2n]/(2n+1)=S(1/√2)=√2ln(1+√2)