Question

數列{

1

n(n+1)

}的前n項和為S_n，則

lim

n→∞

Sn=______．

Answer 1

由題意可得an=

1

n(n+1)

=

1

n

-

1

n+1

∴Sn=

1

1×2

+

1

2×3

+… +

1

n(n+1)

=1-

1

2

+

1

2

-

1

3

+…+

1

n

-

1

n+1

=1-

1

n+1

=

n

n+1

∴

lim

n→∞

Sn=

lim

n→∞

n

n+1

=

lim

n→∞

1

1+ 1 n

=1
故答案為：1