Question

计算

lim

n→∞

[

1

1×3

+

1

2×4

+

1

3×5

+…+

1

n(n+2)

]=

3

4

．

Answer 1

分析：先利用裂项求和可得，

1

1×3

+

1

2×4

+…+

1

n(n+2)

=

3

4

-

2n+3

2(n+1)(n+2)

，代入可求极限

lim

n→∞

[

1

1×3

 +

1

2×4

+…+

1

n(n+2)

]=

lim

n→∞

[

3

4

-

2n+3

2(n+1)(n+2)

]

解答：解：∵2[

1

1×3

+

1

2×4

+…+

1

n(n+2)

]
=1-

1

3

+

1

2

-

1

4

+…+

1

n

-

1

n+2

=1+

1

2

-

1

n+1

-

1

n+2

=

3

2

-

2n+3

(n+1)(n+2)

∴

1

1×3

+

1

2×4

+…+

1

n(n+2)

=

3

4

-

2n+3

2(n+1)(n+2)

∴

lim

n→∞

[

1

1×3

 +

1

2×4

+…+

1

n(n+2)

]=

lim

n→∞

[

3

4

-

2n+3

2(n+1)(n+2)

]=

3

4

故答案为：

3

4

点评：本题主要考查了数列极限的求解，解题的关键是利用裂项求和，但本题裂项是考生容易出现错误的地方，由于

1

(n+2)n

=

1

2

(

1

n

-

1

n+2

)中的

1

2

容易漏掉，注意此类裂项的规律

1

n(n+k)

=

1

k

×(

1

n

-

1

n+k

)