已知数列(An)中,A1=1⼀3,AnA(n-1)=A(n-1)-An(n>=2),数列Bn满足Bn=1⼀An

2025-05-11 18:10:32
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回答1:

由AnA(n-1)=A(n-1)-An两边同时除以AnA(n-1),便得到1/An-1/A(n-1)=1,所以B1=3,Bn-B(n-1)=1,于是Bn=n+2.
所以An=1/(n+2)
则An/n=1/[n(n+2)]=[1/n-1/(n+2)]/2
于是Tn=[1/1-1/3+1/2-1/4+1/3-1/5+……+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2
=[1/1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-[1/(n+1)+1/(n+2)]/2
由于1/(n+1)+1/(n+2)>1/(n+2)+1/(n+2)
所以Tn<3/4-1/(n+2)