Question

設(shè)an=

n

k=1

1

k(n+1-k)

，求證：當(dāng)正整數(shù)n≥2時(shí)，a_n+1＜a_n．

Answer 1

分析：先對(duì)數(shù)列的通項(xiàng)化簡，再作差，證明其大于0，即可證得結(jié)論．

解答：證明：由于

1

k(n+1-k)

=

1

n+1

(

1

k

+

1

n+1-k

)，因此
an=

n

k=1

1

k(n+1-k)

=

n

k=1

1

n+1

(

1

k

+

1

n+1-k

)=

1

n+1

(1+

1

n

+

1

2

+

1

n-1

+…+

1

n

+1)=

2

n+1

(1+

1

2

+…+

1

n

)=

2

n+1

n

k=1

1

k

，
于是，對(duì)任意的正整數(shù)n≥2，有

1

2

(an-an+1)=

1

n+1

n

k=1

1

k

-

1

n+2

n+1

k=1

1

k

=(

1

n+1

-

1

n+2

)

n

k=1

1

k

-

1

(n+1)(n+2)

=

1

(n+1)(n+2)

(

n

k=1

1

k

-1)＞0，即a_n+1＜a_n．

點(diǎn)評(píng)：本題考查數(shù)列與不等式的綜合，考查學(xué)生分析解決問題的能力，對(duì)數(shù)列的通項(xiàng)化簡是關(guān)鍵．