微积分 第4讲 清华大学 高等数学 课件

‡È© A (1)
ƒ[ÿ
14ù
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-‡Ï• e ±n 10 2 F ‘U• ±F 9 29 F
2 / 28
êƤ 5êÆ©z6 /mc/ 5êƆ<©6 /mh/
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1 3 ù£
n n a n→∞
= 0,
?
Œ
¤‡(Ø.
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~ 12. y²: lim
log n α n→∞ n
= 0 (α > 0). ∃N1 > 0 + 1 < eαεn. Œ eαεn , < ε. y.
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y²: ∀ε > 0, du lim ¦ ∀n >
1
n+1 αεn = 0, e n→∞ n+1 N 1 , þk e αεn < 1, = n
n→∞
e lim an = A, lim bn = B , K (1) ∀α, β ∈ R, lim (αan + βbn) = αA + βB ;
n→∞
(2) lim anbn = ( lim an)( lim bn) = AB ;
n→∞ n→∞ n→∞
(3) lim
an n→∞ bn
=
n→∞
lim an
n
|ak − A|
k =N1 +1
M n − N1 ε ε ε + · < + = ε. n n 2 2 2 Ïd¤y(ؤá.
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(: ;.
1 log n n→∞
4•
lБайду номын сангаасm
= 0. •¯!
éê¼ê'~êO•
log n α n→∞ n
lim
= 0 (Ù¥ α > 0). •¯!
˜¼ê'éê¼êO•
m©üNk.. duUCê ÙñÑ5,
džüNk.½n•,¤á.
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‡
~ 1. ∀n an = ¦y: ê 1, ½Â 1 1+ n
êe
½Â
n
, bn =
1 1+ n
n+1
.
{an} Ú {bn} Âñ ‡
Ó˜4•.
5: T4•Ò´Í¶ ê ¢Sþ‰Ñ e
ê e. þãùü‡
þ!e“kn%C”.
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~ 14. y²: lim
1 √ n n→∞ n!
= 0.
ε −n = 0, l n→∞ n! −n ·‚k εn! < 1, •=
y²: ∀ε > 0, dc~Œ• lim ∃N > 0 ¦ ∀n > N , 1 ·‚k √ < ε. dd y. n n! ~ 15. y²: lim
·1
1 1− n+1 (n+1)+1 n+2
=
1
n+1 n+2 n+2
=
n+2 n+1
n+2
= bn+1.
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u ´ ∀n
1, þk a1 ∀n
n+1
an
bn
b1. düN Ù4•
k.½nŒ• {an} Ú {bn} þÂñ. ©O• a, b. 5¿
1 bn = 1 + n
1, ·‚k
n +k
ε = k a
n+k
l=0
ε n+k (a − 1)k+1 k a k+1 εnk+1 (a − 1)k+1 > nk , k (k + 1)!a
•=·‚k
nk an
< ε. dd
y.
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~ 10. y²: lim y²: du α 0<
nα an n[|α|]+1 an ,
nα n n→∞ a
1, - a n =
n k =1
1 . k2
¦y: {an} Âñ.
1 y²: ∀n 1, ·‚k an+1 = an + (n+1) 2 > an , ê {an} 4O. q ∀n 1, ·‚k
n
an =
k =1
1 k2
n
n
1+
k =2
1 k (k − 1)
= 1+
k =2
1 1 1 − = 2 − < 2. k−1 k n
lim bn n→∞
=
A B
(e B = 0).
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£
b ê
: Y%
n
{an}, {bn}, {xn} ÷v: ∀n > n0, þk an xn bn ;
(1) ∃n0 > 0 ¦
n→∞
(2) lim an = lim bn = A;
n→∞
Kê
{xn} Âñ… lim xn = A.
n→∞
y²: - k = [|a|] + 1. K ∀ε > 0, eN = max k,
|a|k+1 k !ε |a|k+1 k !ε ,
, l
@o ∀n > N , ·‚k n >
an n!
=
|a|k+1 n·k !
n− 1 j =k +1
|a| j
|a|k+1 n·k !
< ε,
?
Œ•¤‡(ؤá.
- N = [(N1 + 1) α ] + 1. K ∀n > N , ·‚k nα > N1 + 1, l [nα ] > N1, ?
α] α
nα < [nα ] + 1 < eαε[n u´ α log n < αεnα ,
log n nα
~ 13. y²: lim
an n→∞ n!
= 0 ( a ∈ R) .
ε ∀n > N1, þk |an − A| < 2 . N1
M=
k =1
|ak − A|, N = max N1,
2M ε
.
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K ∀n > N , ·‚k a1 + a2 + · · · + an −A n 1 n
N1
1 n
n
|ak − A|
k =1
k =1
1 |ak − A| + n
î‚4O, e ∀n 4~, e ∀n 1 , an
î‚4~, e ∀n †4~ê
Ü¡üNê
½n 1. (üNk.½n) üN4Okþ. üN4~ke. y²: b an ê (.½n•Tê ∀n > N , an |an − A| < ε, l ê ê 7Âñ; 7Âñ.
{an} üN4Okþ.. Kd kþ(. A, u´ ∀n A < aN + ε
n! n n n→∞
= 0.
n! nn
y²: ∀n > 1, ·‚k 0 < u´dY%
=
1 n
n j =2
j n
1 n.
ná•Œ ¤‡(Ø.
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~ 16. e lim an = A, ¦y:
n→∞
a1 + a2 + · · · + an = A. n→∞ n lim y²: ∀ε > 0, d4•½ÂŒ•, ∃N1 > 0 ¦
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£
lim ( 1 + n − lim √ 3
: ;.~f
√ 3 n) = 0, lim
2n2 +n+2 2 n→∞ n −3
n→∞
= 2. k 0
a0 +···+ak nk n→∞ b0 +···+b n
=
0 ek< , a b e k = ,
•
n→∞ n→∞
ê, ai, bj ∈ R … b = 0.
m
n an k 1
= max ak , Ù¥ ak 1 k m k =1 √ √ lim n n = 1, lim n a = 1 (a > 0). lim
n→∞
0.
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ê
4•
n→∞
OŽ
n•
n+1 n .
~ 7. |^½Ây²: lim xn = 1, Ù¥ óêž, xn =
n− 1 n ,
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y²: ∀n dAÛ²þ
1, d½ÂŒ• 0 uŽâ²þŒ
an
bn . ,
an
1 n ·1 = 1+ n 1 n 1+ n + 1 n+1 n+1 n+1 1 = 1+ n+1 = an+1.
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Ó
/, ·‚•k bn = 1 1+ n
n+1
= 1
1
1 1 − n+1 n+2 n+1
e nlim an = 0 →∞
: ê
4•
5Ÿ (Y)
{bn} k., K nlim an bn = 0. →∞
√
n→∞
e {an} šK…Âñu A, K lim •˜5: eê
an =
√
A.
Âñ, KÙ4••˜.
k•>5: UCk•‘ØUCñÑ5. þ!5: ê Âñ Âñ …= § ?¿f uÑ).