Question

（2006•廣州二模）若Sn=

1

12+2

+

1

22+4

+

1

32+6

+…+

1

n2+2n

（n∈N^*），則

lim

n→∞

Sn=

3

4

．

Answer 1

分析：化簡數(shù)列的通項(xiàng)，利用裂項(xiàng)法，求出數(shù)列的和，然后通過數(shù)列極限的運(yùn)算法則，求出極限．

解答：解：因?yàn)?span id="lttksj3" class="MathJye">

1

n2+2n

=

1

2

 (

1

n

-

1

n+2

)，
所以Sn=

1

12+2

+

1

22+4

+

1

32+6

+…+

1

n2+2n

=

1

2

(

1

1

-

1

3

+

1

2

-

1

4

+

1

3

-

1

5

+…+

1

n

-

1

n+2

)
=

1

2

(1+

1

2

-

1

n+1

-

1

n+2

)．
所以

lim

n→∞

Sn=

lim

n→∞

 

1

2

(1+

1

2

-

1

n+1

-

1

n+2

)=

1

2

(1+

1

2

)=

3

4

．
故答案為：

3

4

．

點(diǎn)評：本題考查數(shù)列通項(xiàng)公式的應(yīng)用，裂項(xiàng)法求法數(shù)列的和，數(shù)列極限的應(yīng)用，考查計(jì)算能力．