問一道數列放縮求和題。求證 1 n n 1 n 20 25其中n屬於正整數

時間 2023-01-25 09:54:16

1樓:匿名使用者

n從1加到k

σ[1/n(n+1)(n+2)]

=σ1/2[(1/n-1/(n+1)]-1/2[1/(n+1)-1/(n+2)]

=σ1/2[(1/n-1/(n+1)]-σ1/2[1/(n+1)-1/(n+2)]

=1/2[1-1/(k+1)]-1/2[1/2-1/(k+2)]

=1/4-1/2(k+1)-1/2(k+2)

<1/4

關鍵在於第一步怎麼把1/n(n+1)(n+2)拆開,這利用到高等代數裡的一個結論,如果你是高中生,可以用待定係數法求解,設1/n(n+1)(n+2)=a/n+b/(n+1)+c/(n+2),兩邊同乘以n,然後令n=0,可得a=1/2,同理可求得b=-1,c=1/2。然後把-1/(n+1)拆成兩項,再結合一下,

即可得到σ1/2[(1/n-1/(n+1)]-σ1/2[1/(n+1)-1/(n+2)]

2樓:理論工作

由於1/(n+1)(n+2)=1/(n+1)-1/(n+2),那麼觀察數列:

1/n(n+1)(n+2)+1/(n+1)((n+2)(n+3)+1/(n+2)(n+3)(n+4)+...

=1/n*[1/(n+1)-1/(n+2)]+1/(n+1)*[(1/n+2)-1/(n+3)]+1/(n+2)*[1/(n+3)-1/(n+4)]-...

=1/n(n+1)-[(1/n-1/(n+1)]*1/(n+2)-[(1/(n+1)-1/(n+2)]*1/(n+3)-[(1/(n+2)-1/(n+3)]*1/(n+4)-...

=1/n(n+1)-[1/n(n+1)(n+2)+1/(n+1)((n+2)(n+3)+1/(n+2)(n+3)(n+4)+...]

即1/n(n+1)(n+2)+1/(n+1)((n+2)(n+3)+1/(n+2)(n+3)(n+4)...=1/2n(n+1)

即原式的極限是1/4=0.25