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