高二证明题 1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)<1

放缩法
n>3时,n!=1*2*...*(n-1)*n>(n-1)n
所以原式
<1/2+1/6+1/(3*4)+1/(4*5)+..+1/((n-1)n)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/(n-1)-1/n
=1-1/n
<1
得证

首先，注意到1/2+1/4+1/8+...=1
我们只要比较这两组和就可以，
对于第n项，(n>2)
(n+1)!=(n+1)n(n-1)..*2*1
2^n =2* 2*2*....*2
显然(n+1)!>2^n
所以1/(n+1)!<1/2^n
所以1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)<1/2+1/4+1/8+...=1
所以1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)<1

因为左边<1/2+1/4+1/8+...+1/2^n=1/2+(1/2-1/4)+(1/4-1/8)+...+(1/2^(n-1)/2^n)=1-1/2^n<1,得证