∵a₁=1，a₂=

1

2

，

2

an

=

1

an+1

+

1

an-1

∴数列{

1

an

}是以1为首项，以d=

1

a2

-

1

a1

=1为公差的等差数列

∴

1

an

=1+n-1=n

∴an=

1

n

∵b_n=a_na_n+1=

1

n(n+1)

=

1

n

-

1

n+1

∴b₁+b₂+…+b_n=1-

1

2

+

1

2

-

1

3

+…+

1

n

-

1

n+1

=1-

1

n+1

=

n

n+1

故答案为：

n

n+1