Question

已知数列an:1/2,1/3+2/3,1/4+2/4+3/4,..,1/10+2/10+3/10+..+9/10,..,

Answer 1

根据题意通项

an=(1+2+3+.+n)/(n+1)

=[(1+n)n/2]/(n+1)

=n/2

裂项：1/(an*an+1)=4/[n(n+1)]=4[1/n-1/(n+1)]

{1/(an*an+1)}的前n项和

Tn=4[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]

=4[1-1/(n+1)]

n-->∞时,Tn-->4

{1/(an*an+1)}的所有项的和为Tn的极限为4