(2) n+1 (n+1)!10n+1 n+1 →>00(n→>∞) u 10 n!10 故级数∑n发散 (3), im+=lim (2n-1)·2n n→)∞nn→0(2n+1)·(2m+2) 1 (2n-1)·2n1 级数∑收敛 n→)0 H=1 故级数22n(2 收敛 n-1)(2) →  (n → ), ! 10 10 ( 1)! 1 1 n n u u n n n n  + = +  + 10 + 1 = n . 10 ! 1 故级数  发散  n= n n (3) (2 1) (2 2) (2 1) 2 lim 1 lim +  + −  = → + → n n n n u u n n n n  = 1, 比值审敛法失效, 改用比较审敛法 , 1 1 级数  2 收敛  n= n  . 2 (2 1) 1 1 故级数  收敛  n= n n − 4 1 1 (2 1) 2 1 2 = −  n n n n →  lim