Question

已知数列｛an｝的通项公式为an=(3n-2)/(3n+1)

已知数列｛an｝的通项公式为an=(3n-2)/(3n+1)
求证：0< an <1

Answer 1

an=(3n+1-3)/(3n+1)
=1-3/(3n+1)

n>=1
3n+1>=4
所以0<1/(3n+1)<=1/4
-3/4<=-3/(3n+1)<0
1/4<=1-3/(3n+1)<1
所以0<an<1

Answer 2

an=(3n+1-3)/(3n+1)
=1-3/(3n+1)

n≥1
3n+1≥4
所以0<1/(3n+1)≤1/4
-3/4≤-3/(3n+1)<0
1/4≤1-3/(3n+1)<1
所以0<an<1
对an， 用分离的方法哦