函数项级数一 -答案

S? ?p nmgdxwjk? xiaojuan yk e jisuanjixueyuanyxy yangxiaoyuan 2015.03.27 1 ?!?SSSNNN 1. n)?Vg; 2. ?CauchyOK; Weierstrass?O!Dirichlet ?O!Abel?O?5; 3. ?5Y5!5 5. 4. ? 2 ?VVVggg( 1. S?5 : x0 I, 0,N(,x0) N,n N : |fn(x) f(x)| 0,N() N,n N,x I : |fn(x) f(x)| 0, sup x0I N(,x0) fn(x) f(x), (n ), x D. limsup nxI |fn(x) f(x)| = 0 S?:?Cauchyn fn(x)3I: x0 I, 0,N(,x0) N,n N,p N: |fn(x0) fn+p(x0)| 0,N() N,n N,p N,x I,: |fn(x) fn+p(x)| 0,N() N,n N,p N,x I : |un+1(x) + + un+p(x)| . ? ?Pun(x)3D?7?un(x)3D u. ? Sn(x) S(x), (n ), x D. limsup nxD |S(x) Sn(x)| = 0. 3. ?5?O (1) Weierstrass?O; (2) Dirichlet?O; (3) Abel?O. 4 SK 1. e?. (1) X n=1 1 1 + xn ; lim n |un+1(x) un(x) | = lim n 1 + xn 1 + xn+1 = (a) 1 |x| 1, X n=1 un(x); (b)1,1 x 1,”?;(c)3,x = 1”?. ?|x| 1,un(x) = 1 1 + xn 9 0(n ),?u; ?x = 1,un(x) = 1 2 9 0(n ),?u; ?x = 1, lim n un(x)3,?u; n,?(,1) (1,+). (2) X n=1 1 n(n + 1)(x 2 + x + 1)n; lim n n p |un(x)| = lim n n s 1 n(n + 1)(x 2 + x + 1)n= x2+ x + 1 ?x2+ x + 1 1=(x + 1)x 0 1 x 1,un(x) 9 0(n ),?u -x = 0,? = X n=1 1 n(n + 1);x = 1,? = X n=1 1 n(n + 1). ? X n=1 un(x)?1,0. 5 2. ?e?3m?5. (1)fn(x) = (1 x)xn, I = 0,1; lim n fn(x) = lim n(1 x)x n = f(x) = 0;(x 0,1) d|fn(x) f(x)| = (1 x)xn9(1 x)xn 0 = xn1n (n + 1)x N?|fn(x) f(x)|3x = n n + 1? l?n= sup x0,1 |fn(x) f(x)| = (1 n n + 1)( n n + 1) n = 1 n+1 (1 + 1 n) n 0(n ) lim n n= 0,fn(x)30,1uf(x) = 0. (2)fn(x) = nx(1 x)n, I = 0,1; lim n fn(x) = lim n nx(1 x)n= f(x) = 0;(x 0,1) n= sup x0,1 |fn(x) f(x)| |fn( 1 n) f( 1 n)| = (1 1 n) n 1 e (n ) lim n n6= 0,fn(x)30,1uf(x) = 0. (3)fn(x) = xn 1 + xn , a.0 x 1 (0 1); lim n fn(x) = lim n xn 1 + xn = f(x) = 0(x 0,1 ) n= sup x0,1 |fn(x) f(x)| = xn 1 + xn (1 )n 0(n ) lim n n= 0,fn(x)30,1 uf(x) = 0 6 b.1 x 1 + (0 0). lim n fn(x) = lim n xn 1 + xn = f(x) = 1(x 1 + ,+) n= sup x0,1 |fn(x) f(x)| = 1 1 + xn 1 (1 + )n 0(n ) lim n n= 0,fn(x)31 + ,+)uf(x) = 0 7 3. ?e?3m?5. (1) X n=1 nx 1 + n5x2 , x (,+); da2+ b2 2ab?:| nx 1 + n5x2 | 1 2n 3 2 (|x| +) dWeierstrass?O,?3(,+) (2) X n=1 (1)n(x + n)n nn+1 , x 0,1; Pan(x) = (1)n n ,bn(x) = (1 + x n) n; (I)? X an(x) (II)z?x 0,1,bn(x)nN4O (III)x 0,1n,k|bn(x)| = (1 + x n) n e,=bn(x)Nk. dAbel?O?30,1 (3) X n=2 ln(1 + x nln2n),l x 0; ln(1 + x nln2n) = x nln2n + o( 1 n2 ) ?|x| l,| x nln2 | l nln2n,? X a nln2n9 X 1 n2 ?3(l,l) (4) X n=1 (1)n+1 x2+ n, x (,+). un(x) = (1)n1 x2+ x,K|un+1