a1=1
an= a(n-1)/[ n(a(n-1) +1) ]
1/an =[ n(a(n-1) +1) ]/a(n-1)
= n + n/a(n-1)
两边除n!
(1/an)/n! = 1/(n-1)! + [1/a(n-1)]/(n-1)!
(1/an)/n! -[1/a(n-1)]/(n-1)! = 1/(n-1)!
(1/an)/n! -[1/a1]/1! = 1/1!+1/2!+...+1/(n-1)!
(1/an)/n! -1 = 1/1!+1/2!+...+1/(n-1)!
(1/an)/n! =1+ 1/1!+1/2!+...+1/(n-1)!
lim(n->无穷)[ 1+1/1!+1/2!+...+1/(n-1)!] =e
=>
lim(n->无穷) [(1/an)/n! ] =e
lim(n->无穷) 1/(n!.an) =e
lim(n->无穷) n!.an =1/e
追问太感谢了
请问能写出来发图片吗
两边除n阶乘的下面第四部是怎么来的呢
追答1/an = n + n/a(n-1)
(1/an)/n! = [ n + n/a(n-1) ]/n!
= n/n! +[n/a(n-1) ]/n!
=1/(n-1)! + [1/a(n-1) ]/(n-1)!