例3判断级数敛散性: n+ (1)∑ n n nt·n n n 解 n (1+ n lim(1+-2)=lim(1+-2)"=e"=1 n→)0 n→)0 limn"=limx =explim-In x =explim-=e=1; x-00例 3 ; ) 1 ( (1) : 1 1 = + + n n n n n nn 判断级数敛散性 解 n n n n n nn n u ) 1 ( 1 + = , ) 1 (1 2 1 n nnn+ = n n n n n n n 1 2 2 ) ] 1 ) lim[(1 1 lim(1 2 + = + → →  1 ; 0 = e = x x n n n x 1 1 lim lim → → = ln } 1 exp{lim x x →  x = } 1 exp{limx→ x = 1 ; 0 = e =