设S1=1+1⼀(1^2)+1⼀(2^2),S2=1+1⼀(2^2)+1⼀(3^2),S3=1+1⼀(3^2)+1⼀(4^2)......Sn=1+1⼀[n^2+1⼀(n+1)^2].

Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
故√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以:
√S1=1+1-1/2
√S2=1+1/2-1/3
√S3=1+1/3-1/4
....
√Sn=1+1/n-1/(n+1)
s= 1+1-1/2 +1+1/2-1/3 1+1/3-1/4 +1+1/(n(n+1)))=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)

因为Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
所以√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以S=3/2+7/6+13/12+...+[n(n+1)+1]/[n(n+1)]=(1+1/2)+(1+1/6)+...+(1+1/(n(n+1)))=n+[1/2+1/6+...+1/(n(n+1))]=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)

1+1/n²+1/(n+1)²通分时分子不用展开应等于[n(n+1)]²+(n+1)²+n² =[n(n+1)]²+2n(n+1)+1 =[n(n+1)+1]²

(n+1)的平方-1/（n+1)

√S1=1+1/(1×2) √S2=1+1/(2×3) ….√Sn=1+1/(n×(n+1))
S=(1+1+…..+1)+1/(1×2)+1/(2×3)+…+1/(n×(n+1))=n+[1-1/(n+1)]
= n+n/(n+1)