Question

計算

lim

n→∞

[

1

1×3

+

1

2×4

+

1

3×5

+…+

1

n(n+2)

]=

3

4

．

Answer 1

分析：先利用裂項求和可得，

1

1×3

+

1

2×4

+…+

1

n(n+2)

=

3

4

-

2n+3

2(n+1)(n+2)

，代入可求極限

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

故答案為：

3

4

點評：本題主要考查了數(shù)列極限的求解，解題的關(guān)鍵是利用裂項求和，但本題裂項是考生容易出現(xiàn)錯誤的地方，由于

1

(n+2)n

=

1

2

(

1

n

-

1

n+2

)中的

1

2

容易漏掉，注意此類裂項的規(guī)律

1

n(n+k)

=

1

k

×(

1

n

-

1

n+k

)