e的两种计算方式
(e=lim_{n o infty}(1+frac{1}{n})^n)
(e=sum_{n=0}^{+infty}frac{1}{n!})
(即，e=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}frac{1}{3!}+cdotcdot)
(所以2<e<1+1+frac{1}{2^2}+frac{1}{2^3}+frac{1}{2^4}+cdotcdotcdot)=3
(即2<e<3)
(可知e不是整数，用反证法，假设e是有理数，即e=frac{p}{q},且q不是1，即qgeqslant2,则)
(q!cdot e=q!sum_{n=0}^{+infty}frac{1}{n!}quadquadquad(1))
(quadquadquad=sum_{n=0}^{+infty}q!frac{1}{n!})
(quadquadquad=sum_{n=0}^{q}q!frac{1}{n!}+sum_{n=q+1}^{+infty}q!frac{1}{n!})
(上式的右侧第二项为：\)
(sum_{n=q+1}^{+infty}q!frac{1}{n!})
(quad=sum_{n=q+1}^{+infty}frac{1}{q+1}+frac{1}{q+1}frac{1}{q+2}+cdotcdot)
(quadleqslantfrac{1}{3}+frac{1}{3^2}+frac{1}{3^3}+cdotcdot<=frac{1}{2})
((1)式的左侧quad q!cdot e=q!frac{p}{q}=(q-1)!p,是整数，而右侧有分数，显然矛盾)