设已知数列为a(n)，则a(n)=1/n(n+1) = (1/n)- ( 1/(n+1) )
s(n)=a(1)+a(2)+a(3)+...+a(n-1)+a(n)=(1-1/2) + (1/2-1/3)+(1/3-1/4)...+( 1/(n-1)-1/n )+( 1/n-1/(n+1) )=1-(1/(n+1))=n/(n+1)(后项和前项相消)
s(1)=1/2，s(2)=2/3,s(3)=3/4
1/n(n+1)=1/n-1/n+1
1/1x2+1/2x3+1/3x4+…+1/n(n+1)
=1-1/2+1/2-1/3+1/3-1/4+.......+1/n-1/n+1
=1-1/n+1
=n/n+1
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