按定义考察部分和Sn=n k=1 (?1)k+1(1 uk +1 uk+1 )=n k=1 (?1)k+11 uk +n k=1 (?1)k+11 uk+1 =?n k=1 (?1)k uk +n+1 l=1 (?1)l1 ul =1 u1 +(?1)n+1 un+1 →1 u1 (n→+∞),所以原级数收敛.再考察取绝对值后的级数∞ n=1 (1 un +1 un+1 ).注意1 un +1 un+1 1 n =n un +n+1 un+1 ?n n+1 →2, ∞ n=1 1 n 发散?∞ n=1 (1 un +1 un+1 )发散.故选:C.