Question

在数列an中，an=1/(n*(n+1)*(n+2))，那么Sn的极限是？

要有详细步骤哟。。
最后1-2+1/2不是应该等于-1/4吗？

Answer 1

首先分析an=1/(n*(n+1)*(n+2))=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
所以Sn=a1+a2+...an
=1/(1*2*3)+1/(2*3*4)+...1/[n(n+1)(n+2)]
=1/2*[1/(1*2)-1/(2*3)]+1/2*[1/(2*3)-1/(3*4)]+..+1/2*[n(n+1)]-1/[(n+1)(n+2)]
=1/2{1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+n(n+1)-1/[(n+1)(n+2)]}
=1/2*[1/2-1/(n^2+3n+2)]
所以limSn=lim{1/2*[1/2-1/(n^2+3n+2)]}=lim(1/4)=1/4

Answer 2

an=1/(n*(n+1)*(n+2))=1/2*(1/n-2/(n+1)+1/(n+2))
2Sn=(1/1-2/2+1/3)+(1/2-2/3+1/4)+(1/3-2/4+1/5)...+(1/n-2/(n+1)+1/(n+2)）
=1/1-2/2+1/2+1/(n+1)-2/(n+1)+1/(n+2)
=1/2-1/(n+1)+1/(n+2) (n>=3)
Sn极限为1/4

Answer 3

-2s(n 1)s(n)=s(n 1)-s(n), -2=1/s(n)-1/s(n 1), 1/s(n 1)=1/s(n) 2, {1/s(n)}是首项为1/s(1)=3,公差为2的等差