在正项等比数列an中,a1＜a4=1,若集合A={n|(a1-1/a1)+(a2-1/a2)+…+(an-1/an)≤0

设公比为q
因a1＜a4=a1*q^3=1
所以0

(a1-1/a1)+(a2-1/a2)+…+(an-1/an)≤0设Sn为等比数列an的前n项和，Tn为正项等比数列1/an的前n项和，上式可化为Sn-Tn<=0,Sn=a1(1-q^n)/(1-q),Tn=1/a1*[1-(1/q)^n]/[1-(1/q)]=(q^n-1)/[a1*q^(n-1)*(q-1)]代入可得[(q^n-1)/(q-1)]*(a1an-1)<=0,即a1an<=1,即a...