Question

已知数列{an}的通项公式an=1/[n*(n+2)],则其前n项和为Sn,limSn=

做出来了加分

Answer 1

∵ an＝1/[n*(n+2)]=1/2[1/n-1/(n+2)]
∴Sn＝a1+a2+a3+a4+a5+。。。。。。+an-2+an-1+an
＝1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+。。。。。。+(1/(n-2)-1/n)+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))]
＝1/2[1+1/2-1/(n+1)-1/(n+2)]
∴limSn=lim{1/2[1+1/2-1/(n+1)-1/(n+2)]}=1/2(1+1/2)=3/4

Answer 2

可得到an=1/2*{[1/n]-[1/(n+2)]}，
观察多个，可利用消元法

追问

步骤

追答

2*Sn=(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+[1/(n-1)]-[1/(n+1)]+[1/n]-[1/(n+2)]
= 1+1/2-[1/(n+1)]-[1/(n+2)]
2*limSn=3/2-lim {[1/(n+1)]+[1/(n+2)]} =3/2
limSn=3/4

Answer 3

裂项求和
1/[n(n+2)]=1/2*[1/n-1/(n+2)]
sn=1/2[1-1/2+1/2-1/3~~1/n]=1/2*(1-1/n)
limsn=1/2

Answer 4

Sn=1/2{(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+[1/(n-1)]-[1/(n+1)]+[1/n]-[1/(n+2)]}
= 1/2{1+1/2-[1/(n+1)]-[1/(n+2)]}（注意这里是间隔两项消去的，所以最后剩下前两项和最后两项）
limSn=3/4