这个不是等比级数吧?
1/[k(k+1)]=1/k - 1/(k+1)
∑1/[k(k+1)] = 1/（1*2）+1/（2*3）+...+1/[n(n+1)]+...
lim k->∞
1-1/2+1/2-1/3+..+1/k-1/(k+1)
=lim k->∞{1-1/(k+1)}
=1