Question

在数列an中,已知a₁=⅔,an+1=2an/an+1,n=1,2,3.一证明数列｛1/an-1

Answer 1

a(n+1)=2an/(1+an)【取倒数】

1/a(n+1)=(1+an)/2an

2/a(n+1)=(1/an)+1

[2/a(n+1)]-2=(1/an)-1

于是[1/a(n+1)]-1=(1/2)[(1/an)-1]

(1/a1)-1=1/2

数列{(1/an)-1}是以1/2为首项,1/2为公比得等比数列

（1/an)-1=(1/2)^n

（2）

1/an=1+(1/2)^n

n/an=n+(n/2^n)

sn=1/a1+2/a2+……+n/an

=1+(1/2)+2+(2/2²）+……+n+(n/2^n)

=（1+2+……+n）+[(1/2)+(2/2²)+(3/2³)……+(n/2^n)]

=[n(n+1)/2]+[(1/2)+(2/2²)+(3/2³)……+(n/2^n)]

令bn=(1/2)+(2/2²)+(3/2³)+……+(n/2^n)

2bn=1+(2/2)+(3/2²)……+[n/2^(n-1)]

bn-2bn=-1-(1/2)-(1/2²)-……-[1/2^(n-1)]+(n/2^n)

-bn=-2+[(n+2)/2^n]

bn=2-[(n+2)/2^n]

于是sn=[n(n+1)/2]+bn

=[n(n+1)/2]+2-[(n+2)/2^n]