∵

1

n(n+1)

=

1

n

-

1

n+1

，∴Sn=(1-

1

2

)+ (

1

2

-

1

3

)+…+（

1

n

-

1

n+1

）=1-

1

n+1

=

n

n+1

∴S _n+1•S_n+2=

n+1

n+2

•

n+2

n+3

=

n+1

n+3

=

3

4

，解得n=5

故答案为：5．