∑ (-1)^n[1-cos(1/n)]对应的正项级数
∑ [1-cos(1/n)] = ∑2 {sin[1/(2n)]}^2
< ∑2 [1/(4n^2)] = (1/2)∑ 1/n^2 ,
后者收敛,则原级数绝对收敛.