bn=n/an,且数列bn前n项记为Sn,求证Sn<2

显示全部楼层 · 2012-6-25 21:37:44

证：设数列{an}公比为q，数列为正项数列，各项均为正，因此q>0。a5/a1=q^4=32/2=16q=-2(<0，舍去)或q=2an=a1q^(n-1)=2×2^(n-1)=2?bn=n/an=n/2?Sn=b1+b2+...+bn=1/2 +2/22+3/23+...+n/2?Sn/2=1/22+2/23+...+(n-1)/2?+n/2^(n+1)Sn-Sn/2=Sn/2=1/2+1/22+1/23+...+1/2?-n/2^(n+1)=(1/2)(1-1/2?)/(1-1/2)...

千问 · 2012-6-25 21:37:44

an=2^nSn=1/2+2/4+3/8............+n/2^nSn/2=0+1/4+2/8..............+n/2^(n+1)
( 乘以公比1/2)相减得到Sn/2=1/2+1/4+............1/2^n-n/2^(n+1)Sn=1+1/2+1/4............1/2^(n-1)-...