解：拆项因为e^x=∑(0,+∞)[1/n!]x^n那么∑(1,+∞)[(2n+1)/n!]x^n=2∑(1,+∞)[1/(n-1)!]x^n+∑(1,+∞)[1/n!]x^n=(2/x)∑(1,+∞)[1/(n-1)!]x^(n-1)+∑(0,+∞)[1/n!]x^n-1=(2/x)e^x+e^x-1
(2e^x)/x+e^x-1
