数学分析试题集锦（一）：10

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?±sZ?xy"+y0 +xy = 0μ[ T3 y = a0 +a1x+a2x2 +￠￠￠+anxn; 5Aμai = 0(i = 1;2; ￠￠￠ ;n): 2
 k??" ?a0;a1;￠￠￠ ;an;￠￠￠ ; P1P n=0 anxn ?@ à?￡Z? (1?x2)y"?2xy0 +l(l +1)y = 0: x2 ? ? ?[[[))) ? ? ?￥￥￥ l l l ? ? ????###'''???ééé 1
p/ ) ?￥?
 (1) 1P n=1 1 (5n?4)(5n+1); (2) 1P n=1 1 4n2?1; (3) 1P n=1 (?1)n?1 2n?1 ; (4) 1P n=1 2n?1 2n ; (5) 1P n=1 rn sinnx;jrj < 1; (6) 1P n=1 rn cosnx;jrj < 1: 2
) ?/ ) ?￥ ???
 (1) 1P n=1 n 2n?1; 1 (2) 1P n=1 ( 12n + 13n); (3) 1P n=1 cos …2n+1; (4) 1P n=1 1 (3n?2)(3n+1); (5) 1P n=1 1p n(n+1)(pn+pn+1): 3
￡ ü? ?10.2. 4
 !) ?1P n=1 unò[ ^?￥
ü) ?￥[üVF?7¤??) ?1P n=1 Un;' Un+1 = ukn+1 +ukn+2 +￠￠￠+ukn+1;n = 0;1;2;￠￠￠ ?k0 = 0;k0 < k1 < k2 < ￠￠￠ < kn < kn+1 < ￠￠￠ :?1P n=1 Un l ?
￡ üe ? ￥) ?9 l ?. x3???[[[))) ? ? ? 1
?Y/ ) ?￥ l ??
 (1) 1P n=1 1p n2+n; (2) 1P n=1 1 (2n?1)22n?1; (3) 1P n=1 n?pn 2n?1 ; (4) 1P n=1 sin …2n; (5) 1P n=1 1 1+an (a > 1); 2 (6) 1P n=1 1 n npn; (7) 1P n=1 ( 12n+1)n; (8) 1P n=1 1 [ln(n+1)]n; (9) 1P n=1 2+(?1)n 2n ; (10) 1P n=1 2n sin …3n; (11) 1P n=1 nn n! ; (12) 1P n=1 nlnn 2n ; (13) 1P n=1 n!2n nn ; (14) 1P n=1 n!3n nn ; (15) 1P n=1 n2 (n+1n)n; (16) 1P n=1 xn (1+x)(1+x2)￠￠￠(1+xn) (x ? 0); (17) 31 + 3￠51￠4 + 3￠5￠71￠4￠7 + 3￠5￠7￠91￠4￠7￠10 +￠￠￠ ; (18) 1P n=1 1 nlnn; (19) 1P n=1 1 (lnn)lnn; (20) 1P n=1 1 2lnn; (21) 1P n=1 1 3lnn; 3 (22) 1P n=1 1 3pn; (23) 1P n=1 n 3pn: 2
 ?¨ ü à T ?íkl ￥¨
V7?Y/ ) ?￥ l ??
 (1) 1P n=1 [e?(1+ 1n)n]p; (2) 1P n=3 lnp cos …n; (3) 1P n=1 (pn+1?pn)p ln n?1n+1; (4) 1P n=1 (pn+a? 4pn2 +n+b): 3
X? ?[) ?1P n=1 un?1P n=1 vn??
 ù1P n=1 max(un;vn)
1P n=1 min(un;vn) ) ?￥ l ????
$ 4
??[) ?1P n=1 an l ?
an+1 ? an(n = 1;2;￠￠￠)
p￡limn!1nan = 0. 5
 ! 8 < : an = 1n2;n 6= k2;k = 1;2;￠￠￠ ; ak2 = 1k2;k = 1;2;￠￠￠ ; p￡:(1) 1P n=1 an l ?; (2) limn!1nan 6= 0: 6
) ?/ ) ?￥ l ??: (1) 1P n=2 1 n(lnn)p; (2) 1P n=2 1 n￠lnn￠lnlnn; 4 (3) 1P n=2 1 n(lnn)1+ lnlnn( > 0); (4) 1P n=2 1 n(lnn)p(lnlnn)q: 7
 ?¨ ?-1?YEù?/ ) ?￥ l ??: (1) 1P n=1 [(2n?1)!!(2n)!! ]p(p ^ L ?); (2) 1P n=1 fi(fi+1)￠￠￠(fi+n?1) n! 1 nfl(fi > 0;fl > 0): 8
 !an > 0;Olimn!1 an+1an = l,p￡limn!1 npan = l.Q- ^?? ?? 9
 ?¨) ? l ?￥A1Hq￡ ü: (1) limn!1 nn(n!)2 = 0; (2) limn!1 (2n)!an! = 0(a > 1): 10
 !an ? 0,O ? fnangμ?,￡ ü) ?1P n=1 a2n l ?. 11
 !?[) ?1P n=1 an l ?,￡ ü1P n=1 pa nan+19 l ?. 12
 !limn!1an = l,p￡: (1)?l > 1 H, +1P n=1 1 nan l ?; (2)?l < 1 H, 1P n=1 1 nan??. ùl = 1 H?μ I 12 ?? x4BBB???[[[))) ? ? ? 1
) ?/ ) ?￥ l ??: 5 (1) 1P n=1 (?1)n pn n+100; (2) 1P n=1 lnn n sin n… 2 ; (3) 1P n=1 (?1)n1+12+￠￠￠+1nn ; (4) 1P n=2 (?1)np n+(?1)n; (5) 1P n=1 sin(…pn2 +1); (6) 1P n=1 (?1)n(n?1)2 3n ; (7) 1P n=1 (?1)n np (p > 0); (8) 1P n=1 1 3n sin n… 2 ; (9) 1P n=1 (?1)ncos2nn ; (10) 1P n=1 (?1)nsin2 nn ; (11) 1P n=1 (?1)n sin xn(x 6= 0); (12) 1P n=1 (?1)nn (n+1)2; (13) 1p2?1 ? 1p2+1 + 1p3?1 ? 1p3+1 +￠￠￠+ 1pn?1 ? 1pn+1 +￠￠￠ ; (14) 1P n=1 (?1)n+1 n a 1+an(a > 0); (15) 1P n=1 sin(n+1n) n ; (16) 1P n=1 sinnsinn2 n : 6 2
) ?/ ) ? ^? ' l ?Hq l ?: (1) 1P n=1 (?1)n n+x ; (2) 1P n=1 sin(2nx) n! ; (3) 1P n=1 sinnx n (0 < x < …); (4) 1P n=1 cosnx np (0 < x < …); (5) 1P n=1 (?1)n np+1n (p > 0); (6) 1P n=2 (?1)n [n+(?1)n]p(p > 0); (7) 1P n=1 (?1)n np+ 1n ; (8) 1P n=1 (?1)n?12n sin2n xn ; (9) 1P n=1 ( xan)n; limn!1an = a > 0; (10) 1P n=1 (?1)nrn+ pn (r > 0); (11) 1P n=1 n!(xn)n; (12) 1P n=1 ln(1+ (?1)nnp ); (13) 1P n=1 (?1)n [pn+(?1)n?1]p; (14) 1P n=1 sin n4… np+sin n4…: 3
 ?¨ O l ?e ??Y/ ) ?￥ ???: 7 (1) a0 +a1q +a2q2 +￠￠￠+anqn +￠￠￠ ;jqj < 1;janj? A (n = 0;1;2;￠￠￠); (2) 1+ 12 ? 13 + 14 + 15 ? 16 +￠￠￠ : 4
p￡:?) ?1P n=1 an(an ? 0) l ?,5) ?1P n=1 a2n l ?.?Q-?? ?,h  è0. 5
?) ?1P n=1 an l ?,Olimn!1 bnan = 1,ù ^???1P n=1 bn9 l ??ù? è 0 an = (?1) n pn ;bn = an + 1n: 6
￡ ü:?) ?1P n=1 an(A)#1P n=1 bn(B)? l ?,O an ? cn ? bn(n = 1;2;￠￠￠) 5) ?1P n=1 cn(C)9 l ?,?) ?(A)D(B)???,ù) ?(C)￥ l ????? 7
￡ ü:?1P n=1 an nx0 l ?,5?x > x0 H, 1P n=1 an nx9 l ?.? 1P n=1 an nx0??,5 ?x < x0 H, 1P n=1 an nx9??. 8
p￡:? ? fnangμK, 1P n=1 n(an ?an?1) l ?,51P n=1 an9 l ?. 9
p￡:?1P n=1 (an ?an?1) ' l ?, 1P n=1 bn l ?,51P n=1 anbn l ?. 10
p￡:?) ?1P n=1 a2n?1P n=1 b2n? l ?,5) ? 1X n=1 janbnj; 1X n=1 ]n +bn)2; 1X n=1 janj n 9 l ?. 11
 !?[ ? fxng??  6Oμ?,p￡: 1X n=1 (1? xnx n+1 ) 8 l ?. 12
 ? fang;fbng,?lSn = nP k=1 ak;￠bk = bk+1 ?bk,p￡: (1)?TfSngμ?, 1P n=1 j￠bnj l ?,Obn ! 0(n !1),51P n=1 anbn l ?,Oμ 1X n=1 anbn = ? 1X n=1 Sn ￠￠bn; (2)?T1P n=1 anD1P n=1 j￠bnj? l ?,51P n=1 anbn l ?. 13
 !1P n=1 an l ?,Olimn!1nan = 0,p￡: 1X n=1 n(an ?an+1) l ?,iO 1X n=1 n(an ?an+1) = 1X n=1 an 14
/  ^d5,￥hóí￡ ü,p￥h Q è: (1)?an > 0,5a1 ?a1 +a2 ?a2 +a3 ?a3 +￠￠￠ l ?; (2)?an ! 0,5a1 ?a1 +a2 ?a2 +a3 ?a3 +￠￠￠ l ?; (3)?1P n=1 an l ?,51P n=1 (?1)nan l ?; (4)?1P n=1 a2n l ?,51P n=1 a3n ' l ?; (5)?1P n=1 an??,5an?t?0; (6)?1P n=1 an l ?,bn ! 1
51P n=1 anbn l ?; (7)?1P n=1 janj l ?, bn ! 1
51P n=1 anbn l ?; 9 (8)?1P n=1 an l ?,51P n=1 a2n l ?; (9)?1P n=1 an l ?,an > 0
5limn!1nan = 0. 15
p/ K(?p > 1) (1) limn!1( 1(n+1)p + 1(n+2)p +￠￠￠+ 1(2n)p); (2) limn!1( 1pn+1 + 1pn+2 +￠￠￠+ 1p2n): x5íííkkk))) ? ? ?DDD}}} ? ? ? ? ? ? 1
?¨ O5,p￡:?T1P n=1 janj,51P n=1 an9 l ?. 2
 !1P n=1 an l ?,p￡:|M #}[?Da ??￥) ? l ?,O μM ]￥? ?. 3
p￡:?) ?1P n=1 (?1)n?1p n×? ?¤￥) ? 1+ 1p3 ? 1p2 + 1p5 + 1p7 ? 1p4 +￠￠￠ ??. 4
￡ ü:?1P n=1 anHq l ?,5 Vü) ?×?, P?) ??s? ? μB 0 ? t_?+1,μB0 ? t_?1. 5
X?Hn = 1 + 12 +￠￠￠+ 1n = c + lnn +rn,c ^x ?è ?, limn!1rn = 0,p ￡: (1) 12 + 14 +￠￠￠+ 12m = 12 lnm+ 12c+ 12rm; (2)?ü) ?1? 12 + 13 ? 14 +￠￠￠￥ò[×?,7 PGQp??[￥BFD GQq?μ[￥BFM?9,5?) ?￥?1ln2+ 12 ln pq. 10 6
p￡:) ?1P n=1 (?1)n+1 n￥üZ( Oe) ^ l ?￥. 7
 7ex = 1P n=0 xn n!,p￡e x+y = ex ￠ey. 8
￡ ü:?) ?￥[F ?|a ??￥) ? l ?,iO]B? ?|= [￥?|M],* 1?? ?|a,N) ?g l ?;i?N I3) ? 1X n=1 (?1)[ pn] n ￥ l ??. 11