Question

数列an是q>1 的等比数列,（a10）^2=a15,Sn=a1+a2+…+an,Tn=1/a1+1/a2+…+1/an

Answer 1

设a0=a,则Sn=a(1-q^n)/(1-q),

Tn=(1/a)[1-(1/q)^n]/[1-(1/q)]

Sn-Tn=[(1-q^n)/(1-q)]*{a-[1/(a*q^(n-1))]},

要使Sn-Tn>0,则a-1/(a*q^(n-1))>0,即a^2>(1/q)^(n-1)

又由(a10)^2=a15得a^2*q^18=a*q^14,

故a=(1/q)^4,由此得q^(n-1)>q^8,由于q>1,故n的最小值是10