Question

设An为数列{(2n-1)/2n}的前n项的积,是否存在实数a,使得不等式An*根号下(2n+1)

Answer 1

注：打字不好.原谅.（1）可设Bn=An*√(2n+1).（n=1,2,3,...).则[B(n+1)]/Bn=[A(n+1)*√(2n+3)]/[An*√(2n+1)]=a(n+1)*√[(2n+3)/(2n+1)]=[(2n+1)/(2n+2)]*√[(2n+3)/(2n+1)]=√[(2n+1)(2n+3)/(2n+2)^2]=√[(4n^2+8n+3)/(4n^2+8n+4)]B(n+1)/BnBn>B(n+1).(n=1,2,3,...).故数列{Bn}是单调递减的非负数列.===>(Bn)max=B1=A1*(√3)=a1*(√3)=(√3)/2.===>(Bn)max=(√3)/2.(2)由题设,应有(√3)/2√33/a-3(1+√3)/2