1/2/（1+1/2）+1/3/(1+1/2)(1+1/3)+.+1/2004/(1+1/2)(1+1/3)(1+1/4)...(1+1/2004)=?

Question

1/2/（1+1/2）+1/3/(1+1/2)(1+1/3)+.+1/2004/(1+1/2)(1+1/3)(1+1/4)...(1+1/2004)=?

Answer 1

各项排列规律：
分子为1/(项数+1),分母为从1+1/2一直连乘到1+1/(项数+1)
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)]
等式左边共有2004-1=2003项.
原式=2(1/2-1/3+1/3-1/4+...+1/2004-1/2005)
=2(1/2-1/2005)
=1-2/2005
=2003/2005