得∑一发散敛该级数的收敛区域为46 六.求下列级数的和 ∑ 2(3n+1)(3n+4(3n+7) 2 左(3k+1)3k+4)(3k+7)1813k+13k+43k+7 1,21121121 2 471071013 3n+13n+43n+7 所以 z(3n+1)(3n+4)(3n+7)24 i n(n+m) 令n充分大,n→ n(n+m) m(nn+m n(n+m)n红k(k+m)mn=如(kk+m m n+1 7+m/5m(1+2+…+1 2n-1 xP<1级数收敛,所以收敛半径为1.当x=±1时都得到交错级数.由 2n-1 莱布尼兹判别法知收敛.所以收敛区域为[-1,1令s(x)=∑(-1” S(x)=>(-yx2=1+x所以s(x=(x)=「 d x= arctan x -1,1 4.∑m(n+1)x 解Im{m(m+1)1x=|xk1收敛当x=±1得∑m(n+1)及∑(-1)"m(n+1都发散得Â • =1 1 n n 发散敛. 该级数的收敛区域为[4, 6). 六. 求下列级数的和: 1. Â • = 0 (3 +1 )(3 + 4 )(3 + 7 ) 1 n n n n 解. Â Â = = ˙ ˚ ˘ Í Î È + + + - + = + + + n k n k k k k k k k 1 1 3 7 1 3 4 2 3 1 1 18 1 (3 1 )(3 4 )(3 7 ) 1 = ˙ ˚ ˘ Í Î È + + + - + - + + - + + - + + + 3 7 1 3 4 2 3 1 1 13 1 10 2 7 1 10 1 7 2 4 1 7 1 4 2 1 18 1 n n n L = 24 1 3 7 1 3 4 1 4 1 1 18 1 ˙ Æ ˚ ˘ Í Î È + + + - - n n . 所以 24 1 (3 1 )(3 4 )(3 7 ) 1 0 = + + + Â • n= n n n 2. Â • = 1 ( + ) 1 n n n m 解. ˜ ¯ ˆ Á Ë Ê + = - n n + m m n n m 1 1 1 ( ) 1 . 令 n 充分大, n Æ • Â Â Â= Æ• = Æ• • = ˜ ¯ ˆ Á Ë Ê + = - + = + n k n n k n n n n m k k m m k k m 1 1 1 1 1 lim 1 ( ) 1 lim ( ) 1 = ˜ ¯ ˆ Á Ë Ê ˜ = + + + ¯ ˆ Á Ë Ê + - - + + + + - m Æ• m n n m m m n 1 2 1 1 1 1 1 1 1 2 1 lim 1 1 L L L 3. Â • = - - - - 1 2 1 1 2 1 ( 1 ) n n n n x 解. | | 1 2 1 | | lim 2 2 1 = < - - Æ• x n x n n n 级数收敛, 所以收敛半径为 1. 当 x = ± 1时都得到交错级数. 由 莱 布尼 兹判 别法知 收敛 . 所 以收敛 区域 为[ - 1, 1].令 s(x ) = Â • = - - - - 1 2 1 1 2 1 ( 1 ) n n n n x . s'(x ) = 2 1 1 2 2 1 1 ( 1 ) x x n n n + Â - = • = - - 所 以 dx x x s x s x dx x x arctan 1 1 ( ) ' ( ) 0 2 0 = + = = Ú Ú , [-1, 1]. 4. Â • = + 1 ( 1 ) n n n n x 解. lim ( + 1) | | = | | < 1 Æ• n n x x n n n 收敛. 当 x = ± 1得Â • = + 1 ( 1 ) n n n 及Â • = - + 1 ( 1 ) ( 1 ) n n n n 都发散