limL=1≠0 根据级数收敛的必要条件,原级数发散 pIn(n+ 2) ∑ (>0 H=1 解lim In(n+ 2) -limiN(n+2), n→+o n→》+o n≥2时,n+2<e",从而有 1<m(n+2)<"n,由于 limn=1, lim/ln(n+2)=1, limu n→+Qlim = 1  0, → n n u 根据级数收敛的必要条件，原级数发散．   =  + + 1 ( 0). ) 1 ( ln( 2) (2) n n a n a n 解 n a n u n n n n n 1 ln( 2) lim lim + + = →+  →+  lim ln( 2), 1 n n n a = + →+ 2 , 2 , n  n  时 n +  e 从而有 1 ln( 2) , n n  n +  n lim = 1, →+ n n 由于 n lim ln( + 2) = 1, →+ n n n . 1 lim a n u n n = →+