对于数列{an},若存在一个常数M,使得对任意的n∈N*,都有|an|≤M,则称{an}为有界数列. (Ⅰ)判断an=2+sinn是否为有界数列并说明理由. (Ⅱ)是否存在正项等比数列{an},使得{an}的前n项和Sn构成的数列{Sn}是有界数列?若存在,求数列{an}的公比q的取值范围;若不存在,请说明理由. (Ⅲ)判断数列an=
|
(Ⅰ)1≤an=2+sinn≤3,
故{an}为有界数列…(2分)
(Ⅱ)设公比为q,当0<q<1时,Sn=
<a1(1-qn) 1-q
,a1 1-q
则正数数列{Sn}满足|Sn|<
,即为有界数列;a1 1-q
当q=1时,Sn=na1→+∞,故为无界数列;
当q>1时,Sn=a1+a2+…+an>na1→+∞,此时为无界数列.
综上:当且仅当0<q<1时,{Sn}为有界数列…(6分).
(Ⅲ){an}为无界数列,事实上an=
+1 3
+1 5
+…+1 7
>1 2n-1
+1 4
+1 6
+…+1 8 1 2n
∴2an>
+1 3
+1 4
+1 5
+…+1 6
+1 2n-1 1 2n
∴2a2n>
+1 3
+1 4
+1 5
+…+1 6
=(1 2•2n
+1 3
)+(1 4
+1 5
+1 6
+1 7
)+(1 8
+…+1 9
)+…+(1 16
+1 2n+1
+…+1 2n+2
)>1 2n+2n
×2+1 4
×4+1 8
×8+…+1 16
×2n=1 2n×2 n 2
∴a2n>n 4
故当n无限增大时an也无限增大,
所以{an}无界…(12分).