證明不等式
|e−(1+1n)n|<3n
證明:
由(二)證明數列{(1+1/n)^(n+1)}爲遞減數列可得
e<(1+1n)n+1=(1+1n)n(1+1n)
可得
e−(1+1n)n<1n(1+1n)n
只要證明
(1+1n)n<3
當n=1時顯然成立
當n>=2時,由於
(1+1n)n=c0n+c1n1n+c2n1n2+...+ckn1nk+...+cnn1nn
因爲
ckn1nk=n(n−1)...(n−k+1)k!·1nk
ckn1nk=1k!·nn·n−1n·n−2n...n−k+1n<1k!<=12k−1;(2<=k<=n)
所以
(1+1n)n<1+12+121+...+12n−1=1+(1−12n)1−12
(1+1n)n<3−12n−1<3
所以
|e−(1+1n)n|<3n