(1) s(n-1)=a(n)-2
s(n)=a(n+1)-2
兩式相減
s(n)-s(n-1)=a(n)=a(n+1)-a(n)
2a(n)=a(n+1)為等比數(shù)列.因為a(1)=2,從而a(n)=2^n
(2) b(n)=1/log(2^n)=1/(nlog2)
故t(n)=1/log2[1/(n+1)+1/(n+2)+……+1/(2n)]
因為[1/(n+2)+1/(n+3)+……+1/(2n+2)]-[1/(n+1)+1/(n+2)+……+1/(2n)]
=1/(2n+1)-1/(2n+2)=1/[(2n+1)(2n+2)]>0
所以t(n)為遞增數(shù)列.即t(1)=1/2 < t(2)=1/3+1/4=7/12