a1=1/1,
a2=(1+2)/2,
a3=(1+2+3)/3,
a4=(1+2+3+4)/4,
...
a100=(1+2+3+4+...+100)/100,
所以
an=(1+2+3+...+n)/n,
又1+2+3+4+...+n=n(1+n)/2,
所以an=n(1+n)/2n=(1+n)/2,
Bn=1/(An(An+1))=1/[(1+n)^2/4+(1+n)/2],
Bn=1/[(n^2+2n+1+2n+2)/4]=4/(n+3)(n+1),
Bn=2*[1/(n+1)-1/(n+3)]
Sn=B1+B2+...+Bn=Bn+B(n-1)+B(n-2)+B(n-3)+...+B4+B3+B2+B1,
Sn=2*{[1/(n+1)-1/(n+3)]+[1/n-1/(n+2)]+[1/(n-1)-1/(n+1)]+[1/(n-2)-1/n]+...+[1/5-1/7]+[1/4-1/6]+[1/3-1/5]+[1/2-1/4]},
中間對(duì)消后得,Sn=2*{[-1/(n+3)]-1/(n+2)]+1/3+1/2}=5/3-2*[1/(n+3)+1/(n+2)]=5/3-(4n+10)/(n^2+5n+6),
Cn=(1/（2n）)An=(1/2n)*(1+n)/2=(1+n)/4n,
Tn=[1+n(1+n)/2]/[4*n(1+n)/2]=[2+n(1+n)]/n(1+n)=2/n(1+n)+1