求幂级数∞∑∞n-0(2n+1)x2n的收敛域及和函数.

解 收敛半径R=1 x=+-1时，级数发散，故收敛域（-1,1）∫想0s(x)dx=∞∑n-0)∫x 0(2n+1)x2ndx∅=∞∑n-0 x2n+1=x/1-x2 s(x)=1+x2/(1-x2 )2