求极限题目

更新时间2019-12-04 01:50:27

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