∑1/n(n+1) = 1/(1*2) + 1/(2*3) + 1/(3*4) + .... + 1/(n(n+1)) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .... + (1/n - 1/(n+1) )去掉括号,除了第一项和最后一项抵消 = 1 - 1/(n+1)n->∞, 1/(n+1) ->0lim(n->∞) ∑1/n(n+1) = 1