第2个回答 2017-10-24
级数收敛,就是其前n项和的极限存在。
由于 1/[n(n+1)(n+2)]=1/2*{1/[n(n+1)]-1/[(n+1)(n+2)]} ,因此
Sn=1/(1*2*3)+1/(2*3*4)+......+1/[n(n+1)(n+2)]
=1/2*{[1/(1*2)-1/(2*3)]+[1/(2*3)-1/(3*4)]+......+1/[n(n+1)]-1/[(n+1)(n+2)]}
=1/2*{1/2-1/[(n+1)(n+2)]}
=1/4-1/[2n(n+1)(n+2)] ,
当 n 趋于无穷时,Sn→1/4 ,
因此,级数收敛本回答被网友采纳