∑[x^2n\(2n-1)]=x∑[x^(2n-1)\(2n-1)] (把x提出来了)设g(x)=∑[x^(2n-1)\(2n-1)]一阶导数g'(x)=∑x^(2n-2)=1/(1-x^2)g(x)=积分1/(1-x^2)=1/2 ln|(1+x)/(1-x)|则∞∑n=1[x^2n\(2n-1)]=xg(x)=x/2 ln|(1+x)/(1-x)|...