公差为1的等差数列 A1=1 A2=2

已知数列{an}满足:a1=1,a2=2,且{bn}是公差为1的等差数列. 设bn=an/(n+2)! ,数列{bn}的前n项和为Sn, 求证:1/6≤ Sn <1/2
打错了...... 不好意思 且{a(n+1)/an}是公差为1的等差数列.....

a<1>=1,a<2>=2, =====> a<2>/a<1> = 2
{a<n+1> / a<n>} 是公差为1的等差数列 ======> a<3>/a<2> = 2+1 = 3, a<3> = 3a<2> = 6
a<n+1> / a<n> = (n + 1), n>=2
a<n+1> = a<n> * (n+1)

a<1>=1,a<2>=2= 2!, a<3> = a<2> * 3 = 2! * 3 = 3!, a<n> = n!

b<n> = n! / (n+2)! > 0
b<1> = 1 / (1+2)! = 1/3! = 1/6 =========> Sn >= 1/6 ------------- （1）

Sn = 1/3! + 2!/4! + 3!/5! + 4!/6! + ....+ (n-1)!/(n+1)! + n!/(n+2)!
= 1/(2*3) + 1/(3*4) + 1/(4*5) + .....+ 1/[n(n+1)] + 1/[(n+1)(n+2)]
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ..... + 1/n - 1/(n+1) + 1/(n+1) - 1/(n+2)
= 1/2 - 1/(n+2)
< 1/2 ---------------- （1）

由（1）（2）得
1/6 <= Sn < 1/2

a(n+1)/an=n+1
那么an=n!
bn=an/(n+2)!=1/(n+1)(n+2)

则Sn=1/2-1/(n+2)

则Sn=1/2-1/(n+2)>=1/2-1/3=1/6