由最后一项化简整理得:
[﹙n+1﹚²+1]/[﹙n+1﹚²-1]
=[n²+2n+2]/[n﹙n+2﹚]
=[n﹙n+2﹚+2]/[n﹙n+2﹚]
=1+2/[n﹙n+2﹚]
=1+[1/n-1/﹙n+2﹚]
∴以上每一项都可以拆成这种形式:
Sn=[1+﹙1/1-1/3﹚]+[1+﹙1/2-1/4﹚]+[1+﹙1/3-1/5﹚]+[1+﹙1/4-1/6﹚]+……+﹛1+[1/﹙n-1﹚-1/﹙n+1﹚]﹜+﹛1+[1/n-1/﹙n+2﹚]﹜
=1×n+﹛[1+1/2]+[-1/﹙n+1﹚-1/﹙n+2﹚]﹜
=n+n/﹙n+1﹚+n/[2﹙n+2﹚]