1⼀2⼀1+1⼀2+1⼀3⼀（1+1⼀2）*（1+1⼀3)....一直加到分子为1⼀1999

考察一般项第n项：
[1/(n+1)]/[(1+1/2)(1+1/3)...(1+1/(n+1)]
=[1/(n+1)]/[(3/2)(4/3)...(n+2)/(n+1)]
=[1/(n+1)]/{[3×4×...×(n+2)]/[2×3×...×(n+1)]}
=[1/(n+1)]/[(n+2)/2]
=2/[(n+1)(n+2)]
=2[1/(n+1)-1/(n+2)]

(1/2)/(1+1/2)+(1/3)/[(1+1/2)(1+1/3)]+...+(1/1999)/[(1+1/2)(1+1/3)...(1+1/1999)]
=2(1/2-1/3+1/3-1/4+...+1/2000-1/2001)
=2(1/2-1/2001)
=1999/2001

这位大侠的一般项公式是对的，可是最终结果不对！应该是
(1/2)/(1+1/2)+(1/3)/[(1+1/2)(1+1/3)]+...+(1/1999)/[(1+1/2)(1+1/3)...(1+1/1999)]
=2(1/2-1/3+1/3-1/4+...+1/1999-1/2000)
=2(1/2-1/2000)
=999/1000