（1）b1=

1

a1

=

1

P

∴bn+1-bn=

1 an+1 - 1 an = 1+an an - 1 an =1&

故数列{b_n}是以b1=

1

P

为首项，以1为等差的等差数列              

（2）证明：bn=

1

an

=

1

p

+(n-1)⇒an=

1

n+ 1 p -1

∵0＜p＜1

∴ 1 p -1＞0

⇒an=

1

n+ 1 p -1

＜

1

n

a1

2

+

a2

3

+

a3

4

+…+

an

n+1

＜

1

2×1

+

1

3×2

+

1

4×3

+…+

1

(n+1)n

=1-

1

2

+

1

2

-

1

3

+

1

3

-

1

4

+…+

1

n

-

1

n+1

=1-

1

n+1

＜1