求极限题目
请问第8题和第10题怎么解？
8、因为(n-1)/n!=1/(n-1)!-1/n!，则∑(k:2 ……n)(k-1)/k!=∑(k:2 ……n)[1/(k-1)!-1/k!]
=(1-1/2!)+(1/2!-1/3!)+……+(1/(n-2)!-1/(n-1)!)+(1/(n-1)!-1/n!)
=1+(-1/2!+1/2!)+(-1/3!+1/3!)+……+(-1/(n-1)!+1/(n-1)!)-1/n!
=1-1/n!
所以，其极限为1。
10、这是无穷乘积的极限问题，可通过对数变换转化为无穷和的极限。
(1-2/2*3)(1-2/3*4)……(1-2/n*(n+1))=exp{ln[(1-2/2*3)(1-2/3*4)……(1-2/n*(n+1))]}
=exp{ln(1-2/2*3)+ln(1-2/3*4)+……+ln(1-2/n*(n+1))}
当n趋向于无穷大时，2/n*(n+1)趋向于0。由等价无穷小替换公式：ln(1-x)~-x，故有
ln(1-2/n*(n+1))~-2/n*(n+1)=-2（1/n-1/(n+1)）
所以
ln(1-2/2*3)+ln(1-2/3*4)+……+ln(1-2/n*(n+1))~-2[1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=-1+2/(n+1)
n趋向于无穷大时，极限为-1，从而原极限为e^-1=1/e。
8】n→∞
lim[1/2!+2/3!+3/4!+……+(n-1)/n!]
=lim{(1-1/2!)+(1/2!-1/3!)+(1/3!-1/4!)+……+[1/(n-1)!-1/n!]}
=lim(1-1/n!)
=1-0
=0
10】n→∞
lim{[1-2/(2*3)][1-2/(3*4)][1-2/(4*5)]*……*[1-(2/n)/(n+1)]}
=lim{[(1*4)/(2*3)][(2*5)/(3*4)][(3*6)/(4*5)]*……*[(n-1)(n+2)/(n*<n+1>)]}
=lim{[1/2*2/3*3/4*……*(n-1)/n]*[4/3*5/4*6/5*……*(n+2)/(n+1)]}
=lim[1/n*(n+2)/3]
=(1/3)lim(1+2/n)
=1/3*(1+0)
=1/3
lim(1/2！+2/3！+3/4！+（n-1）/n！）
=lim（1/2+1/2-1/3！+1/3！-1/4！+.....-1/(n-1)！+1/(n-1)！-1/n！)
=lim(1-1/n！)
=1
1-2/[n(n+1)]=([n(n+1)]-2)/[n(n+1)]=(n^2+n-2)/[n(n+1)]=[(n-1)(n+2)]/[n(n+1)]
∴lim(n→∞)[(1-1/3)(1-1/6)．．．．(1-2/[n(n+1)]
=lim(n→∞)(1·4)/(2·3)·(2·5)/(3·4)·(3·6)/(4·5)·[(n-1)(n+2)]/[n(n+1)]
=lim(n→∞)[1·2·3*...(n-1)·(4·5·....(n+2)]/(2·3·.....·n·3·4·...·(n+1)
=lim(n→∞)[1·(n+2)/3n]
=1/3