裂项求和:当公差为零时,显然成立.
当公差不为零时,因为1/anan+1=1/d*[1/an-1/an+1]
1/a1a2+1/a3a4+…+1/anan+1=1/d*[1/a1--1/an+1]=1/d*[(an+1-an)/a1an+1]
=1/d*[nd/a1an+1]=n/a1an+1