Question

（2012•南充三模）已知S_n是数列{

1

n2+3n+2

}的前n项和，则

lim

n→∞

Sn等于（　　）

• A．1
• B．

  1

  2

• C．

  1

  6

• D．

  5

  6

Answer 1

分析：由

1

n2+3n+2

=

1

(n+1)(n+2)

=

1

n+1

-

1

n+2

，利用裂项可求S_n，进而可求极限

解答：解：∵

1

n2+3n+2

=

1

(n+1)(n+2)

=

1

n+1

-

1

n+2

∴Sn=

1

2

-

1

3

+

1

3

-

1

4

+…+

1

n+1

-

1

n+2

=

1

2

-

1

n+2

lim

n→∞

Sn=

lim

n→∞

(

1

2

-

1

n+2

)=

1

2

故选B

点评：本题主要考查了裂项求解数列的和及数列极限的求解，属于基础试题