从而原级数不绝对收敛;但它却是满足莱布尼兹条件 的交错级数,即 limu =lim =0且Ln a+on a+b(n+1) n+1 n1→0 n-y00a+bn 则原级数条件收敛 (3∑(-1)"1(m2-1) 解因un=n-1>0,而 LInn_1 li n→0 Nn-I= lim-1 (1-Inn) lim ● n→0 n→0 n n 设∫(x)=x2=ex I21-Inx f(x)=e x9 1 1 1 (3) ( 1) ( 1) n n n n  − =  − − 从而原级数不绝对收敛； 1 1 1 ( 1) u u n n a bn a b n =  = + + + + 则原级数条件收敛. 1 1 0, n 解 因 而 u n n = −  1 ln 1 lim 1 n n n e n → − 2 2 1 (1 ln ) limn 1 n n n → − − = =  − 1 ln ( ) x x x f x x e = = ln 2 1 ln ( ) x x x f x e x − 设  =   但它却是满足莱布尼兹条件 的交错级数, 即 1 lim lim 0 n n n u → → a bn = = + 且 1 lim 1 n n n n →  − =