(9)√n+1-√hi (10)1+2+3+ 证明: (1)对ve>0,由 于m+-叫-m+<第一2,要使m+-<只要<即可 n+1 取N=|2+1,则当n>N时,当x <e总成立,所以士→0(n→ (2)对ve>0,由于 <e,只要<e即可。取N=+1, 则当n>N时 <E总成立,所以=→0(n→∞) (3)对ve>0,由于 n+(-1 要使/n+(-1)2 01<即可。取N[+1,则当n>N时,+(-1)-0<e总成立,所以+(-=1 要 (4)对ve>0,由 要使 <ε,只要-<ε即可。取N 当n>N时,1-0<总成立,所以一0(m→∞) (5)设Sn= 1)2+11 对v>0,由于Sn=2(a-3+ 2+3-…+(-1)+1 设6n=1 当n=2k+1时,有0<bn=1-(-3)-(7-2) (2k-2k+1)<1:当n=2时,有0<6n=1-(2-3)-(4-3)-…-(-2-2k-1)-2k<1 总之,和0<6n<1从而Sn-0=Sn=5<要使Sn-0|<,只要士<即可,取N=[1+1 则当n>N时,|Sn-0<总成立,所以1-1+1 +(-1)6m+1→0(n→∞) (6)对ve>0,由于n>mnn,则e”>n,于是e-n<,从 0|=-+e-n 使(-1)”(0990-0<e,只要(0.99<c即可。取N=2500m+1,则当n>N时,(-1)(0990 0<E总成立,所以(-1)”(0.99902→0(n→∞) ()对v>0,由于+c--0=1<1,要使1+--0<e,只要2<即可,取N= ,则当n>N时,+--0<总成立,所以1+c-一0m→x) (8)对ve>0,由于e-n<e=1,则 要使 <ε,只要-<ε即可 取N=|+1.,则当n>N时, <总成立,所以一→0(m→∞) ()对ve>0,由于vm+一-则=m+1+分F 2V示,要使m+1-V-0<,只要2 e即可。取N=/1 2+1,则当n>N时,m+一一0<c总成立,所以+1-√一m (10)对v>0,由于1+2+3+…+-=2+、∠mn21n 3 要使 <E,只要-<ε即可 取N=1+1.,则当n>N时,1+2+3++n <e总成立,所以1+2+3+…+n
16 (8) e −n n (9) √ n + 1 − √ n (10) 1 + 2 + 3 + · · · + n n3 y²: (1) È∀ε > 0ßdu n + 1 n2 + 1 − 0 = n + 1 n2 + 1 < 2n n2 = 2 n ßᶠn + 1 n2 + 1 − 0 < εßêá 2 n < ε=å" N = 2 ε + 1, Kn > Nûß n + 1 n2 + 1 − 0 < εo§·ß§± n + 1 n2 + 1 → 0(n → ∞) (2) È∀ε > 0ßdu sin n n − 0 = sin n n 6 1 n ßᶠsin n n − 0 < εßêá 1 n < ε=å"N = 1 ε + 1ß Kn > Nûß sin n n − 0 < εo§·ß§± sin n n → 0(n → ∞) (3) È∀ε > 0ßdu n + (−1)n n2 − 1 − 0 = n + (−1)n n2 − 1 < n + 1 n2 − 1 = 1 n − 1 ßᶠn + (−1)n n2 − 1 − 0 < εßê á 1 n − 1 < ε=å"N = 1 ε + 1ßKn > Nûß n + (−1)n n2 − 1 − 0 < εo§·ß§± n + (−1)n n2 − 1 → 0(n → ∞) (4) È∀ε > 0ßdu 1 n! − 0 = 1 n! < 1 n ßᶠ1 n! − 0 < εßêá 1 n < ε=å"N = 1 ε + 1ßK n > Nûß 1 n! − 0 < εo§·ß§± 1 n! → 0(n → ∞) (5) Sn = 1 n − 1 2n + 1 3n − · · · + (−1)n+1 1 n2 È∀ε > 0ßduSn = 1 n (1 − 1 2 + 1 3 − · · · + (−1)n+1 1 n ) δn = 1− 1 2 + 1 3 −· · ·+(−1)n+1 1 n ßKSn = δn n n = 2k+1ûßk0 < δn = 1−( 1 2 − 1 3 )−( 1 4 − 1 5 )−· · ·− ( 1 2k − 1 2k + 1 ) < 1¶n = 2kûßk0 < δn = 1−( 1 2 − 1 3 )−( 1 4 − 1 5 )−· · · − ( 1 2k − 2 − 1 2k − 1 )− 1 2k < 1" oÉßk0 < δn < 1 l |Sn−0| = Sn = δn n < 1 n á¶|Sn−0| < εßêá 1 n < ε=å"N = 1 ε +1ß Kn > Nûß|Sn − 0| < εo§·ß§± 1 n − 1 2n + 1 3n − · · · + (−1)6n + 1 1 n2 → 0(n → ∞) (6) È∀ε > 0ßdun > ln nßKe n > nßu¥e −n < 1 n ßl 1 n + e −n − 0 = 1 n + e−n < 2 n ßá ¶|(−1)n (0.999)n−0| < εßêá(0.999)n < ε=å"N = 2500 ln 1 ε +1ßKn > Nûß|(−1)n (0.999)n− 0| < εo§·ß§±(−1)n (0.999)n → 0(n → ∞) (7) È∀ε > 0ßdu 1 n + e −n − 0 = 1 n! < 1 n ßᶠ1 n + e −n − 0 < εßêá 2 n < ε=å"N = 2 ε + 1ßKn > Nûß 1 n + e −n − 0 < εo§·ß§± 1 n + e −n → 0(n → ∞) (8) È∀ε > 0ßdue −n < e0 = 1ßK e −n n − 0 = e −n n < 1 n ßᶠe −n n − 0 < εßêá 1 n < ε=å" N = 1 ε + 1ßKn > Nûß e −n n − 0 < εo§·ß§± e −n n → 0(n → ∞) (9) È∀ε > 0ßdu| √ n + 1− √ n−0| = 1 √ n + 1 + √ n < 1 2 √ n ßá¶| √ n + 1− √ n−0| < εßêá 1 2 √ n < ε=å"N = 1 4ε 2 + 1ßKn > Nûß| √ n + 1 − √ n − 0| < εo§·ß§± √ n + 1 − √ n → 0(n → ∞) (10) È∀ε > 0ßdu 1 + 2 + 3 + · · · + n n3 − 0 = n + 1 2n2 < 2n 2n2 = 1 n ßᶠ1 n − 0 < εßêá 1 n < ε=å" N = 1 ε + 1ßKn > Nûß 1 + 2 + 3 + · · · + n n3 − 0 < εo§·ß§± 1 + 2 + 3 + · · · + n n3 → 0(n → ∞)
3.举例说明下列关于无穷小量的定义是错误的 (1)对任意e>0,存在N,当n>N时,成立xn<E (2)对任意e>0,存在无限多个xn,使|rn|< (1)例如:数列{-1+(-1)+1}(或{-n})即{0,-2,0,-2,…}(或{-1,-2,-3,…}满足上述条件,但不是 无穷小量 (2)例如:数列{.2,1,3…,n…}满足上述条件,但不是无穷小量 4.按定义证明 (2)in(0.9-9)=1 (4)xn= 1·223 (n-1) 1(n→∞) 5)inmn=1,此处rn={n4 当n为奇数 当n=3k(k=1,2,3,…) (6) lim rn=1,此处rn= 1+n 当n=3k+ n+n 证明 (1)对ve>0,由于3n2+n_3{_2n+3 4(n+1) m2-1-2=4m2-2<4+1)(-1)= (n≥2),要 3m2+n3 要n工<即可,取N=m(]+12.则当>Nw时,图+-引 <e总成立,所 ()对v>0.由于-1=01)=m,要使的8-1<=只要m<c即可,取N +1,则当n>N时,1.99-1<e总成立,所以0.99…9→1(n→∞) (3)对ve>0,由于ym2+n =2+n+n<2 1,要使+n-1|<e,只 n 要1<c即可。取N 1,则当n>N时 总成立,所以n2+n (4)对ve>0,由于 则n-11=-,要使|xn-1 只要 <c即可。取N=|2+1,则当n>N时,|n-1<总成立,所以xr→1(m→∞) (2>0.由于=1=|m主2-1-要h-1<,只要<即可,取N=+则 总成立,所以 1 2 ()对v>0,由于一3=0,+1-3=n,+2-3=3-+n=n+n+√+6 n=,要使n-<,只要<组且<即可,取N=ms(目 当n>N时,|rn-3<ε总成立,所以rn→3(n→∞)
17 3. fi~`²e'uð˛½¬¥Üÿµ (1) È?øε > 0ß3Nßn > Nûߧ·xn < ε¶ (2) È?øε > 0ß3ÃÅıáxn߶|xn| < ε. ): (1) ~XµÍ{−1 + (−1)n+1}(½{−n})={0, −2, 0, −2, · · · } (½{−1, −2, −3, · · · })˜v˛„^áßÿ¥ ð˛¶ (2) ~XµÍ{1, 1 2 , 1, 1 3 , · · · , 1, 1 n , · · · }˜v˛„^áßÿ¥Ã°˛" 4. U½¬y²µ (1) limn→∞ 3n 2 + n 2n2 − 1 = 3 2 (2) limn→∞ (0. n z }| { 99 · · · 9) = 1 (3) limn→∞ √ n2 + n n = 1 (4) xn = 1 1 · 2 + 1 2 · 3 + · · · + 1 (n − 1) · n → 1(n → ∞) (5) limn→∞ rn = 1ßd? rn = n − 1 n nèÛÍ n + 1 n nè¤Í (6) limn→∞ rn = 1ßd? rn = 3 n = 3k(k = 1, 2, 3, · · ·) 3n + 1 n n = 3k + 1 2 + 1 + n 3 − √ n + n n = 3k + 2 y²µ (1) È∀ε > 0ßdu 3n 2 + n 2n2 − 1 − 3 2 = 2n + 3 4n2 − 2 < 4(n + 1) 4(n + 1)(n − 1) = 1 n − 1 (n > 2)ßᶠ3n 2 + n 2n2 − 1 − 3 2 < εßêá 1 n − 1 < ε=å"N = max( 1 ε + 1, 2), Kn > Nûß 3n 2 + n 2n2 − 1 − 3 2 < εo§·ß§ ± 3n 2 + n 2n2 − 1 → 3 2 (n → ∞) (2) È∀ε > 0ßdu 0. n z }| { 99 · · · 9 −1 = (0.1)n = 1 10n ßᶠ0. n z }| { 99 · · · 9 −1 < εßêá 1 10n < ε=å"N = lg 1 ε + 1, Kn > Nûß 0. n z }| { 99 · · · 9 −1 < εo§·ß§±0. n z }| { 99 · · · 9 → 1(n → ∞) (3) È∀ε > 0ßdu √ n2 + n n − 1 = √ n2 + n − n n =< 1 √ n2 + n + n < 1 2n ßᶠ√ n2 + n n − 1 < εßê á 1 2n < ε=å"N = 1 2ε + 1, Kn > Nûß √ n2 + n n − 1 < εo§·ß§± √ n2 + n n → 1(n → ∞) (4) È∀ε > 0ßduxn = 1 − 1 2 + 1 2 − 1 3 + · · · + 1 n − 1 − 1 n = 1 − 1 n ßK|xn − 1| = 1 n ßá¶|xn − 1| < εß êá 1 n < ε=å"N = 1 ε + 1, Kn > Nûß|xn − 1| < εo§·ß§±xn → 1(n → ∞) (5) È∀ε > 0ßdu|rn − 1| = n ± 1 n − 1 = 1 n ßá¶|rn − 1| < εßêá 1 n < ε=å"N = 1 ε + 1, K n > Nûß|rn − 1| < εo§·ß§±rn → 1(n → ∞) (6) È∀ε > 0ßdu|r3k − 3| = 0, |r3k+1 − 3| = 1 n , |r3k+2 − 3| = √ n − 2 3 − √ n + n = n − 4 n √ n + n + √ n + 6 < n n √ n = 1 √ n ßá¶|rn − 3| < εßêá 1 n < εÖ 1 √ n < ε=å"N = max 1 ε + 1, 1 ε 2 + 1 , K n > Nûß|rn − 3| < εo§·ß§±rn → 3(n → ∞)
5.(1)按定义证明,若an→a(n→∞),则对任意自然数k,an+k→a(n→∞) (2)按定义证明,若an→a(n→∞),则an→|a又反之是否成立? (3)若an|→0,试问an→a是否一定成立?为什么? (1)由于an→a(n→∞),故对v>0,丑N∈2+,当n>N时,|an-al<E,则对vk∈z+,n+k N时,|an+k-叫<E,于是对v>0.彐N∈Z+,当n+k>N时,|an+k-a<,从而an+k a(a △此结论说明:去掉数列的前面有限项,也不影响其收敛性 (2)()由于an→a,故对ve>0,N∈2+,当n>N时,{an-a<e叉anl-l<lan-a,于是 对ve>0,N∈Z+,当n>N时,|an|-回<c成立,即an|→la(n→∞ 反之不 (a)不成立:an=(-1)2,则an|→1,而an无极限 (b)成立:an=,则an|→0,an→0 (3)由于nl→0,故对v>0,N∈2+,当n>N时,|an|-0<e,又an-0=|an-0,于 是对v>0,丑N∈2+,当n>N时,lan <e成立,即an→0n→∞)。从而若lan|→0 则an→0一定成立 6.按定义证明,若xn→a,且a>b,则存在N,当n>N时,成立rn>b 证明:由于xn→a,故对ve>0,丑N∈z+,当n>N时,|xn-0<E,即a-E<xn<a+E.又a>b 故a-b>0,则取E=a-b>0,从而N∈2+,当n>N时,有xn>a-E=a-(a-b)=b即存在N, n>N时,成立xn>b 7.若{xnyn}收敛,能否断定{xn},{yn}亦收敛 敛。故若{xnyn}收敛,不能断定{xn},{m}亦收就1(m=1,2,…),则{xnyn}收敛,但{n},{vm}均 8.利用极限性质及计算证明 (1)lim (n+1)2 1 (2)lim /2 (3)利用(1+b)=∑c=1+m+2(mn=2+…+h )lin=0(a>1) (i)lim n5 证明 ()对w2,和0≤+(+12+…+(2nm2,且 0,则lim n+1)2 ()∈2,有n+1如+、+2++2+n<五 1且lim (3)(i)设a=1+h(h>0),由于0< (1+h)n n-1)h2 为定值,n1→0(n-∞),则 (n-1)2→0从而im=0
18 5. (1) U½¬y²ßean → a(n → ∞)ßKÈ?øg,Íkßan+k → a(n → ∞) (2) U½¬y²ßean → a(n → ∞)ßK|an| → |a|.qáÉ¥ƒ§·º (3) e|an| → 0ߣØan → a¥ƒò½§·ºèüoº y²µ (1) duan → a(n → ∞)ßÈ∀ε > 0, ∃N ∈ Z +ßn > Nûß|an − a| < εßKÈ∀k ∈ Z +, n + k > Nûß|an+k − a| < εßu¥È∀ε > 0, ∃N ∈ Z +ßn + k > Nûß|an+k − a| < εßl an+k → a(n → ∞) 4 d(ÿ`²µKÍc°kÅëßèÿK蟬Ò5" (2) (i) duan → aßÈ∀ε > 0, ∃N ∈ Z +ßn > Nûß|an − a| < ε.q |an| − |a| < |an − a|ßu¥ È∀ε > 0, ∃N ∈ Z +ßn > Nûß |an| − |a| < ε§·ß=|an| → |a|(n → ∞ (ii) áÉÿò½§·" ~µ (a) ÿ§·µan = (−1)nßK|an| → 1ß anÃ4Ŷ (b) §·µan = 1 n ßK|an| → 0, an → 0 (3) du|an| → 0ßÈ∀ε > 0, ∃N ∈ Z +ßn > Nûß |an| − 0 < εßq|an − 0| = |an| − 0 ßu ¥È∀ε > 0, ∃N ∈ Z +ßn > Nûß|an − 0| < ε§·ß=an → 0(n → ∞)"l e|an| → 0ß Kan → 0ò½§·" 6. U½¬y²ßexn → aßÖa > bßK3Nßn > Nûߧ·xn > b. y²µduxn → aßÈ∀ε > 0, ∃N ∈ Z +ßn > Nûß|xn − 0| < εß=a − ε < xn < a + ε.qa > bß a − b > 0ßKε = a − b > 0ßl ∃N ∈ Z +ßn > Nûßkxn > a − ε = a − (a − b) = b.=3Nß n > Nûߧ·xn > b. 7. e{xnyn}¬ÒßUƒ‰½{xn}, {yn}½¬Ò. )µÿU" ~µxn = (−1)n , yn = (−1)n (n = 1, 2, · · ·), xnyn ≡ 1(n = 1, 2, · · ·)ßK{xnyn}¬Òß{xn}, {yn}˛ÿ¬ Ò"e{xnyn}¬ÒßÿU‰½{xn}, {yn}½¬Ò. 8. |^4Å5ü9Oéy²µ (1) limn→∞ 1 n2 + 1 (n + 1)2 + · · · + 1 (2n) 2 = 0 (2) limn→∞ 1 √ n2 + 1 + 1 √ n2 + 2 + · · · + 1 √ n2 + n = 1 (3) |^(1 + h) n = Pn k=0 C k nh k = 1 + nh + n(n − 1) 2 h 2 + · · · + h n y²µ (i) limn→∞ n a n = 0(a > 1) (ii) limn→∞ n 5 e n = 0(e ≈ 2.7) y²µ (1) È∀n ∈ Z +ßk0 6 1 n2 + 1 (n + 1)2 +· · ·+ 1 (2n) 2 6 n + 1 n2 ßÖ limn→∞ n + 1 n2 = 0ßK limn→∞ 1 n2 + 1 (n + 1)2 + · · · + 1 (2n) 2 = 0 (2) È∀n ∈ Z +ßk n n + 1 < 1 √ n2 + 1 + 1 √ n2 + 2 + · · · + 1 √ n2 + n < n n = 1Ö limn→∞ n n + 1 = 1ß K limn→∞ 1 √ n2 + 1 + 1 √ n2 + 2 + · · · + 1 √ n2 + n = 1 (3) (i) a = 1 + h(h > 0)ßdu0 < n a n = n (1 + h) n = n 1 + nh + n(n1) 2 h 2 + · · · + h n < n n(n − 1) 2 h 2 = 2 (n − 1)h2 ßq 2 h2 è½äß 1 n − 1 → 0(n → ∞)ßK 2 (n − 1)h2 → 0.l limn→∞ n a n = 0
(i)设e=1+h(h≈1.7),由0、5 (1+h)n 1+nh+C2h2+.+h" Ch he 720n5 →0(n→∞),则 720n n-5)%6,又2为定值 0(n→∞),从而lim 9.求下列极限 (1)lm3n3+2n2-n+1 (2)inn3+n2+2 (3)lim 1 cos n 1+-+…+ (4) 1四1 0)=wa(=+)-(+++m) n3+2n2-n+13 (1)lim (2)1mxn3+n2+2=0 (3)由于 1a-,故-2一m),刘m,从而(1-)n=0 1+5+…+2n (4)lim 4-2 (5)由于{sn为有界数列,()-01-1,2n2+1→2m→∞), 2⊥1 故lm(imn)( 1 2n2+1 1.2+2.3+…+ (6)im(=2)+3=1m (x)"+1 10.若xn→a>0,试证: (1)√xn (2)√ down+a1xm-1+…+am-1xn+am→√aoam+a1am-1+…+am-1a+am +am-la+am>0) 证明 (1)由于xn→a>0,故对ve>0.N∈z+,当n>N时,|xn-a|<√ae,且√an-va ase,即对上述>0,3N∈z+,当n>N时,|√an-√a<e,从
19 (ii) e = 1 + h(h ≈ 1.7)ßdu0 < n 5 e n = n 5 (1 + h) n = n 5 1 + nh + C2 nh2 + · · · + hn < n 5 C6 nh6 < 720n 5 (n − 5)6h6 ßq 720 h6 è½äß n 5 (n − 5)6 → 0(n → ∞)ßK 720n 5 (n − 5)6h6 → 0(n → ∞)ßl limn→∞ n 5 e n = 0 9. ¶e4ŵ (1) limn→∞ 3n 3 + 2n 2 − n + 1 2n3 − 3n2 + 2 (2) limn→∞ 6n 2 − n + 1 n3 + n2 + 2 (3) limn→∞ 1 − 1 √n 2 cos n (4) limn→∞ 1 + 1 2 + · · · + 1 2 n 1 + 1 4 + · · · + 1 4 n (5) limn→∞ " (sin n!) n − 1 n2 + 110 − 1 1 · 2 + 1 2 · 3 + · · · + 1 (n − 1) · n 2n 2 + 1 n2 − 1 # (6) limn→∞ (−2)n + 3n (−2)n+1 + 3n+1 )µ (1) limn→∞ 3n 3 + 2n 2 − n + 1 2n3 − 3n2 + 2 = 3 2 (2) limn→∞ 6n 2 − n + 1 n3 + n2 + 2 = 0 (3) du √n 2 → 1(n → ∞)ß1 − √n 2 → 0(n → ∞)ßq| cos n| 6 1ßl limn→∞ 1 − 1 √n 2 cos n = 0 (4) limn→∞ 1 + 1 2 + · · · + 1 2 n 1 + 1 4 + · · · + 1 4 n = limn→∞ 1 − ( 1 2 ) n+1 1 − 1 2 1 − ( 1 4 ) n+1 1 − 1 4 = 2 4 3 = 3 2 (5) du{sin n!}èk.Íß n − 1 n2 + 110 → 0, 1 − 1 n → 1, 2n 2 + 1 n2 + 1 → 2(n → ∞)ß limn→∞ " (sin n!) n − 1 n2 + 110 − 1 1 · 2 + 1 2 · 3 + · · · + 1 (n − 1) · n 2n 2 + 1 n2 − 1 # = −2 (6) limn→∞ (−2)n + 3n (−2)n+1 + 3n+1 = limn→∞ ( −2 3 ) n + 1 (−2)(−2 3 ) n + 3 = 1 3 10. exn → a > 0ߣyµ (1) √ xn → √ a (2) p a0xmn + a1x m−1 n + · · · + am−1xn + am → p a0am + a1am−1 + · · · + am−1a + am (Ÿ•a0a m + a1a m−1 + · · · + am−1a + am > 0) y²µ (1) duxn → a > 0ßÈ∀ε > 0, ∃N ∈ Z +ßn > Nûß|xn − a| < √ aεßÖ| √ xn − √ a| = xn − a √ xn + √ a < |xn − a| √ a < εß=È˛„ε > 0, ∃N ∈ Z +ßn > Nûß| √ xn − √ a| < εßl √ xn → √ a(n → ∞)
(2)由于xn→a(n→∞),故aoxm+a1xm-1+…+am-1xn+am→a0am+a1am-1+…+am-1a+am>0, 则据(1)得 aor+alIn +am-1xn+am→√aoam+a1am-1+…+am-1a+am 11.对数列{xn},若x2k→a(k→∞),x2k+1→a(k→∞),证明:xn→a(n→∞) 证明:ve>0,因x2→a(n→∞),故K1∈2+,使当k>K1时,|x2k-a<E成i 又因x2k+1→a(n→∞),故K2∈Z+,使当k>K2时,|x2k+1-a<成立 取N=max{2K1,2K2+1},则当n>N时,若n为偶数,n=2k>N≥2K1,k>K1,|rn-a=|x2k-a|< 若n为奇数,n=2k+1>N≥2K2+1,k>K2,xn-叫=|x2k+1-a<E, 因此rn→a( 12.利用单调有界必有极限,证明 lin a存在,并求出它 (2)xo=1,x1=1+z0 证明 (1)显然x1<x2,假设xn-1<xn,则xn=√2xn-1<√2xn,由归纳法,知{xn}是单调增加的,又xn √2xn-1,故得x=2xn1≤2xn,于是xn≤2,即{xn}由上界。从而imxn存在,记 lim an=l, 在x2=2xn-1两边令n→∞,得2=21,解之得l=2,即 lim a= (2)显然xn≥1,有条件知xn=1+1+xn-12-1 1+xn-1 <2,故{xn}有界。又x1=1+ 3 1 1+1+1=2>1=x0,假设xn1<xm,则n=2 1+xn-1 =xn+1,由 归纳法,知{xn}是单调增加的。从而 lim n存在,记 lim n=l,在xn=2 两边令 即mx=i+即P=1+,解得=1+ n→∞,得l=2-_1 y5,n=1-5(不合题意,舍去) 13.若2=a>0.=b>0a<6,xm+1=√,+=,证明:lmx=lm 证明:由于√an≤红十如且此式相等当且仅当xn=vm,故xm+1≤如n+1等号成立当且仅当xn= n又0<a<b,故x1<v,则由递推公式,得xn+1<mn+1且xn>0,n>0(n∈2+)而xn+1=√nyn nn=n;孙+1≈xn+y 2<y+=孙,则xn<xn+1<孙+1<孙m又由x1=a>0,y=b>0 得a<n<xn+1<+1<<b,说明{xn}与{y}都是单调有界数列,从而{xn},{vm}均有极限 设lmxn=a,limn=B,又由xn+1=V/nvn,得x2+1=xnn,在等式两边令n→∞,得a2=aB又 由0<a<xn<xn+1,得定有0<a≤a,从而a=B即有 lim n= lim yn 14.利用单调有界必有极限证明以下数列必有极限 1 (2) (3)xn=an(a>1,k为正整数) 证明 (1)由于xn+1-n=10,故xn+1>xn,则{xn}为单调增加的又1<mn<1+言+…+ =1 <2,故{xn}有界,于是{xn}存在极限 n(n+1) (2)由于x+1-xn=3+1+1>0,故xn+1>x,则{xn}为单调增加的,又4<xm<+3+…+< 1+…×、13=1,故{xn}有界,于是{xn}存在极限
20 (2) duxn → a(n → ∞)ßa0x m n +a1x m−1 n +· · ·+am−1xn+am → a0a m+a1a m−1+· · ·+am−1a+am > 0ß K p ‚(1) a0xmn + a1x m−1 n + · · · + am−1xn + am → p a0am + a1am−1 + · · · + am−1a + am 11. ÈÍ{xn}ßex2k → a(k → ∞), x2k+1 → a(k → ∞)ßy²µxn → a(n → ∞) y²µ∀ε > 0ßœx2k → a(n → ∞)ß∃K1 ∈ Z +߶k > K1ûß|x2k − a| < ε§·" qœx2k+1 → a(n → ∞)ß∃K2 ∈ Z +߶k > K2ûß|x2k+1 − a| < ε§·" N = max {2K1, 2K2 + 1}ßKn > NûßenèÛÍßn = 2k > N > 2K1, k > K1, |xn−a| = |x2k−a| < εß enè¤Íßn = 2k + 1 > N > 2K2 + 1, k > K2, |xn − a| = |x2k+1 − a| < εß œdxn → a(n → ∞) 12. |^¸Nk.7k4Åßy² limn→∞ xn3ßø¶—ßµ (1) x1 = √ 2, · · · , xn = √ 2xn−1 (2) x0 = 1, x1 = 1 + x0 1 + x0 , · · · , xn+1 = 1 + xn 1 + xn y²µ (1) w,x1 < x2ßbxn−1 < xnßKxn = √ 2xn−1 < √ 2xnßd8B{ß{xn}¥¸NO\ßqxn = √ 2xn−1ßx 2 n = 2xn1 6 2xnßu¥xn 6 2ß={xn}d˛."l limn→∞ xn3ßP limn→∞ xn = lß 3x 2 n = 2xn−1¸>-n → ∞ßl 2 = 2lß)Él = 2ß= limn→∞ xn = 2" (2) w,xn > 1ßk^áxn = 1 + xn−1 1 + xn−1 = 2 − 1 1 + xn−1 < 2ß{xn}k."qx1 = 1 + x0 1 + x0 = 1 + 1 1 + 1 = 3 2 > 1 = x0ßbxn1 < xnßKxn = 2 − 1 1 + xn−1 < 2 − 1 1 + xn = xn+1ßd 8B{ß{xn}¥¸NO\"l limn→∞ xn3ßP limn→∞ xn = lß3xn = 2 − 1 1 + xn−1 ¸>- n → ∞ßl = 2 − 1 1 + l ß=l 2 = 1 + lß)l1 = 1 + √ 5 2 , l2 = 1 − √ 5 2 £ÿ‹Køß§ß = limn→∞ xn = 1 + √ 5 2 " 13. ex1 = a > 0, y1 = b > 0(a < b), xn+1 = √xnyn, yn+1 = xn + yn 2 ßy²µ limn→∞ xn = limn→∞ yn. y²µdu√xnyn 6 xn + yn 2 Öd™É Ö=xn = ynßxn+1 6 yn+1 “§·Ö=xn = yn.q0 < a < bßx1 < y1ßKd4Ì˙™ßxn+1 < yn+1Öxn > 0, yn > 0(n ∈ Z +). xn+1 = √xnyn > √ xnxn = xn, yn+1 = xn + yn 2 < yn + yn 2 = ynßKxn < xn+1 < yn+1 < yn.qdx1 = a > 0, y1 = b > 0ß a < xn < xn+1 < yn+1 < yn < bß`²{xn}Ü{yn}—¥¸Nk.Íßl {xn}, {yn}˛k4Åß limn→∞ xn = α, limn→∞ yn = βßqdxn+1 = √xnynßx 2 n+1 = xnynß3 ™¸>-n → ∞ßα 2 = αβ q d0 < a < xn < xn+1ß½k0 < a 6 αßl α = β=k limn→∞ xn = limn→∞ yn. 14. |^¸Nk.7k4Åy²±eÍ7k4ŵ (1) xn = 1 + 1 2 2 + · · · + 1 n2 (2) xn = 1 3 + 1 + 1 3 2 + 1 + · · · + 1 3 n + 1 (3) xn = n k a n (a > 1, kèÍ) (4) xn = √n a (0 < a < 1) y²µ (1) duxn+1 − xn = 1 (n + 1)2 > 0ßxn+1 > xnßK{xn}è¸NO\. q1 < xn < 1 + 1 12˙ + · · · + 1 n ˙ (n + 1) = 1 + 1 − 1 2 + · · · + 1 n1 − 1 n = 2 − 1 n < 2ß{xn}k.ßu¥{xn}34Å" (2) duxn+1−xn = 1 3 n+1 + 1 > 0ßxn+1 > xnßK{xn}è¸NO\. q 1 4 < xn < 1 4 + 1 3 2 +· · ·+ 1 3 n < 1 3 + 1 3 2 + · · · + 1 3 n = 1 3 1 − 1 3 = 1 2 ß{xn}k.ßu¥{xn}34Å