（1）由已知得a_n+1a_n=a_n-a_n+1a_n

两边同除以a_n+1a_n得出

1

an+1

-

1

an

=1，

∴数列{

1

an

}为公差为1的等差数列，且首项为

1

a1

=2

根据等差数列的通项公式可得

1 an =2+(n-1)=n+1 ∴an= 1 n+1

（2）证明：∵

an

n+1

=

1

(n+1)2

＜

1

n

-

1

n+1

∴

a1 2 + a2 3 + a3 4 +…+ an n+1 ＜ 1 2×1 + 1 3×2 + 1 4×3 +…+ 1 (n+1)n =1- 1 2 + 1 2 - 1 3 + 1 3 - 1 4 +…+ 1 n - 1 n+1 =1- 1 n+1 ＜1