已知数列Sn=2An+（-1)^nn大于一试证明对于任意m大于4有1/A4+1/A5+1/A6+.+1/Am小于7/8我算出An=[2^(n-1)-2(-1)^n]/3

Question

已知数列Sn=2An+（-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8
我算出An=[2^(n-1)-2(-1)^n]/3

Answer 1

An=[2^(n-1)-2(-1)^n]/3 =(2/3)[2^(n-2)+(-1)^(n-1)].
A4=2.
若m为偶数,则
1/A4 +1/A5 +1/A6 +...+1/Am
=1/A4 +(1/A5+1/A6)+...+[1/A(m-1)+1/Am]
其中1/A(m-1)+1/Am
＝(3/2){1/[2^(m-3)+1]+1/[2^(m-2)-1]}
＝(3/2){[2^(m-3)+1+2^(m-2)-1]/[2^(m-3)+1][2^(m-2)-1]}
＝(3/2){[2^(m-3)+2^(m-2)]/[2^(2m-5)-2^(m-3)+2^(m-2)-1]
＜(3/2){[2^(m-3)+2^(m-2)]/[2^(2m-5)][∵-2^(m-3)+2^(m-2)-1＞0]
=(3/2)[1/2^(m-3)+1/2^(m-2)]
∴1/A4 +1/A5 +1/A6 +...+1/Am
=1/A4 +(1/A5+1/A6)+...+(1/A(m-1)+1/Am)
＜(1/2)+(3/2)[1/2^3+1/2^4+1/2^5+...+1/2^(m-2)]
＝(1/2)+(3/2)(1/4)[1-1/2^(m-4)]
＜(1/2)+(3/8)
=7/8.
当m是奇数时,m+1是偶数,所以
1/A4 +1/A5 +1/A6 +...+1/Am
＜1/A4 +1/A5 +1/A6 +...+1/Am+1/A(m+1)
＜7/8.
综上,命题得证.