1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1)
put x =e^(1/n)
1+e^(1/n) +..+e^(1) = [(e^(1/n))^(n+1) -1]/(e^(1/n)-1)
再问： i=1,从1开始取值，为什么不是e^1/n+...+e^(1) .为什么前面要算上1
再答： divided [0,1] into n equal interval with width 1/n ∫(0->1) e^x dx =lim(n->无穷)summation(i:1->n) (1/n) e^(i/n) =lim(n->无穷) [ 1/n [ e^(1/n) + e^(2/n)+.... +e^(n/n) ] consider 1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1) 1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1) put x =e^(1/n) 1+e^(1/n) +..+e^(1)= [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) e^(1/n) +..+e^(1) = [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) -1 ∫(0->1) e^x dx =lim(n->无穷) [ 1/n [ e^(1/n) + e^(2/n)+.... +e^(n/n) ] =lim(n->无穷) [ 1/n [ [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) -1 ] =lim(n->无穷) [ 1/n [ [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) ] - lim(n->无穷)(1/n) = e-1
