Question

数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,

Answer 1

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

即b(n+1)=bn+n+1/2

bn=b(n-1)+(n-1)+1/2

...

b2=b1+1+1/2

所以bn=(bn-b(n-1))+...+(b2-b1)+b1

=((n-1)+1/2)+...+(1+1/2)+1

=(1+2+...+(n-1))+(1/2+1/2+...+1/2)+1

=n(n-1)/2+(n-1)/2+1

=(n^2+1)/2

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

cn=(n^2+2n+2)/[n(n+1)2^(n+2)]=1/2{(n+1)/(n*2^n)-(n+2)/[(n+1)2^(n+1)]}

所以Sn=1/2[2/(1*2^1)-3/(2*2^2)]+1/2[3/(2*2^2)-4/(3*2^3)]+...+1/2{(n+1)/(n*2^n)-(n+2)/[(n+1)2^(n+1)]}

=1/2{2/(1*2^1)-(n+2)/[(n+1)2^(n+1)]}（裂项相消法）

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

首先Sn=1/2-(n+2)/[(n+1)2^(n+2)]0,即Sn单增

所以Sn≥S1=5/16