∴Sn=(1•a+2•a2+3•a3+…+n•an)lga.
∴aSn=(1•a2+2•a3+3•a4+…+n•an+1)lga.
以上两式相减得(1-a)Sn=(a+a2+a3+…+an-n•an+1)lga=[
a(1−an) |
1−a |
∵a≠1,∴Sn=
alga |
(1−a)2 |
(Ⅱ)由bn<bn+1⇒nlga•an<(n+1)lga•an+1⇒lga•an[n-(n+1)a]<0,
∵an>0,
∴lga[n(a-1)+a]>0.①
(1)若a>1,则lga>0,n(a-1)+a>0,故a>1时,不等式①成立;
(2)若0<a<1,则lga<0,不等式①成立⇔n(a-1)+a<0,∴0<a<
n |
n+1 |
综合(1)、(2)得a的取值范围为a>1或0<a<
n |
n+1 |