m之a-2a k=1 故有 m∑a ≤lim∑laxb k=1 n→0 k=1 即 k=1 推论4.1设an=n+iBn,n=1,2,….则级数 ∑a绝对 n= 收敛的充要条件是级数∑α和∑Bn都绝对收敛1 1 k k k k a a   = =    1 1 lim k k n k k a a   → = =  = 故有 1 1 lim lim , k k n n k k a a   → → = =    1 1 k k k k a a   = = 即    推论 4.1 设 . 则级数 绝对 收敛的充要条件是级数 和 都绝对收敛. i , 1,2, n n n a n = + =   1 n n a  =  1 n n   =  1 n n   = 