问一道数列放缩求和题。求证Σ(1⼀n(n+1)(n+2))<0.25其中n属于正整数

n从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

关键在于第一步怎么把1/n(n+1)(n+2)拆开，这利用到高等代数里的一个结论，如果你是高中生，可以用待定系数法求解，设1/n(n+1)(n+2)=a/n+b/(n+1)+c/(n+2)，两边同乘以n，然后令n=0，可得a=1/2，同理可求得b=-1,c=1/2。然后把-1/(n+1)拆成两项，再结合一下，
即可得到Σ1/2[(1/n-1/(n+1)]-Σ1/2[1/(n+1)-1/(n+2)]

由于1/(n+1)(n+2)=1/(n+1)-1/(n+2),那么观察数列:
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)
即原式的极限是1/4=0.25