《高数》积分判别法(最新整理)

1、1积分判别法 若在1,)上 f 减, 非负, 则f (n)收敛 f 收敛.此时 f f (n) f + f (1).11证 2 f f (1) = f (1), 3 f f (2) 2 f , , n+1 f f (n) nf , 相加得n+1 fn1n21nn-11 f (k ) 1k -1f + f (1). 令 n得证.注. 条件可改为 x 充分大时 f 减, 非负.例 1(p 级数) 1 当且仅当 p 1 时收敛.n p证一. p 0 时用积分判别法; p0 时由必要条件.证二p1 时由 n-pn-1 得发散, p1 时用积分判别法.*证三 p1 时由 n-pn-1 得发散. p 1

2、时按下列方法加括号: 括号内的项数依次为 1, 2, 4, 8, 16, , 则 由 1 + 1 2 = 1 , 1 + l + 1 4 = 1 , 及 比2 p3 p2 p2 p-1 4 p7 p4 p4 p-1较判别法知加括号后的级数收敛, 故 p 级数也收敛. n=2 1,n ln p n1, . n ln n ln p nn=3nnqn an备考. 设 f (x) = (x ln px)-1 (x2), 则 p0 时显然 f 减. 而 p 0 时对充分大的 x, f 仍减p 0 时 f (x) = - (x ln px)-2 ln p-1x (ln x + p) e -p), 故可直接

3、应用积分判别法得 (n ln pn)-1 当 p 1 时收敛, p1 时发散.n( 1 ) . n ( 1 ) . n p( 1 ) .sin 1( 1 ).n an(1 + n)3n 2n - 1(1 + n) p+qnn 1 (= 1 0, 或 1 0) ( an+1 = a , 或= a 0). n ! ( an+1 1 或 1n n !上( n !ann + 1nnanee册 p.40.4(5). nn( an+1 = (1 + 1 )n(n + 1) 2 e 1 或 an ). 1( an ).ln nln nnn -1(ln n) pn -1 ln n (p1 时 an ,发散;

4、p1 时取 q 使 pq1, 则 an 0 或 ann-q, 收p敛).nn -1n -qn a(- 1) (a 1) (由 a x - 1 ln a (x0)知 n a - 1 = o( 1 ). p.16.1 (9)类似).* nn-1xn(nn-1 1 ). * 1( (ln ln x) 2 0(x(2n 2+ ln n + 1)n+12(2n 2 )n+12n2(ln n)ln ln nln x), n 充分大时(ln n) ln ln n= exp(ln ln n)2 0, an 收敛, 则 an 与an an+1 收敛. 与an 比较. n(ln n) p (ln ln n) qa

5、n + 1例 3(p.16.9(4). 考察1的收敛性.n=3解 设 f (x) = x (ln x)p (ln ln x) q, 则 f (x) = ln p-1 x (ln ln x) q-1(ln x + p) ln ln x + q ), x充分大时p, q , f (x) 0, 故可用积分判别法.i = dx = du. p1 时取 r使 pr1,3 f (x)ln 3 u p ln q u由 u r1 0 知 i 收敛. p=1 时 i = dt , 当且仅当 q1 时收敛. p 1 或 p=1, q 1 时收敛, 在其它情形发散.*例 4 (p.16.10) an, 非负, 则a

6、n 收敛2m a2m (=2a2 + 4a4 + )收敛.证 设 2 m a2m = s. 因 为 s2n = a1 + a2 + + a2n = a1 + (a2 + a3 ) + (a4 + + a7 )+ +( a2n-1 + l + a2n -1 ) + a2n a1 + 2a2 + 4a4 + = a1 + s, 故n sn s2n a1 + s, 由收敛原理得an 收敛.设a= s, 则由 a a+ a , 2 a a+ a 等得 1n2m a (a+ a ) + (a+ a )n212+434m12342m=12(a 5 + + a 8 ) + = s. 因此2m a2m 的部分

7、和有界, 从而收敛. 应用: 1 收敛2m 1 =2(1-p)m 收敛21-p 1 .n p2mp1收敛 2m1= 1收敛 1 收敛p1.n ln p n2m ln p 2mm p ln p 2m p*例 5 (raabe 判别法) 若 lim n (1 - an+1 ) = l , 则 l 1 时an 收敛, l 1 时 取 p 使 lp1, 则 n 充 分 大 时 n(1- an+1 )p, an+1 1-p (1 - 1 ) p =n - p.anannn(n - 1) - p由比较法, 收敛. l1 时对充分大的 n 有 n (1 - an+1 )1- 1 =n -1anann(n -

8、 1) -1“”“”at the end, xiao bian gives you a passage. minand once said, people who learn to learn are very happy people. in every wonderful life, learning is an eternal theme. as a professional clerical and teaching position, i understand the importance of continuous learning, life is diligent, nothing can be gained, only continuous learning can achieve better self. only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. t