Question

若a_n=1+2+3+…+n，则S_n为数列{

1

an

}的前n项和，则S_n=______．

Answer 1

由题意可得，an=

n(n+1)

2

∴

1

an

=

2

n(n+1)

=2(

1

n

-

1

n+1

)
∴S_n=a₁+a₂+…+a_n
=2(1-

1

2

+

1

2

-

1

3

+…+

1

n

-

1

n+1

)
=2(1-

1

n+1

)=

2n

n+1

故答案为：

2n

n+1