数列极限二

则称{an}是单调增加数列;
如果满足 an+1 ≤ an , n = 1,2,3,
则称{an}是单调减少数列.
单调增加数列与单调减少数列统称为单调数列
例:
an
=
n ,n n +1
= 1,2,3,,是单调增加数列.
例: an
=
1 n
,n
= 1,2,3,,是单调减少数列.
2018/9/27
Edited by Lin Guojian
−a
< ε ,即lim n→∞
xn
=
a.
2018/9/27
Edited by Lin Guojian
1
例: 证明lni→m∞
1+ n2 +1
1 ++ n2 + 2
1 n2 +
n
=
1
证: n = 1 + 1 ++ 1
n2 + n n2 + n n2 + n
n2 + n
< 1 + 1 ++ 1
n2 +1 n2 + 2
2
例 : 设xn
=
n
3n
+ 5n ,求 lim n→∞
xn .
证 :由于n 5n < n 3n + 5n < n 5n + 5n ,
⇒ 5 < n 3n + 5n < n 2 ⋅5n = n 2 ⋅ n 5n = 5⋅ n 2
由于 lim n n→∞
2
=
1,
故由夹逼定理知
:
lim
n→∞
xn
= lim n n→∞
事实上 : (数学归纳法)当n = 1时, 有1+ x ≥ 1+ x成立.
设n = k时, (1+ x)k ≥ 1+ kx成立.
那么当n = k +1时, (1+ x)k+1 = (1+ x)k (1+ x) ≥ (1+ kx)(1+ x) = 1+ kx + x + kx2 = 1+ (k +1)x + kx2 ≥ 1+ (k +1)x.
+
1
n
=
n
+
2
n
⋅
n
+
2
n+1 n+1
n +1n
n
n
n
=
n(n + 2) (n +1)2
n
⋅
n+2 n +1
=
1 −
(n
1 + 1) 2
n
⋅
n+2 n +1
≥
1 −
(n
n + 1) 2
⋅
n+2 n +1
=
n2 + n +1 (n +1)2
⋅
n+2 n +1
=
(n2
7
定义2.2 : 如果存在常数M > 0,使得数列{an}满足 an ≤ M , n = 1,2,,
则称{an }为有界数列, 否则称{an }为无界数列;
例: 设an = (−1)n ,则 an ≤ 1,故{an}为有界数列.
例: 设an = n,则数列{an}为无界数列.
例: 设an = en ,则数列{an}为无界数列.
2018/9/27
Edited by Lin Guojian
12
故对任意的正整数n, 有
2
=
1 +
1 1
≤
1 +
1
n
≤
1 +
1
n+1
≤
1
+
1
2
=
22
=
4
1 n n 1
从而1
+
1 n
n
单调增加且有界,
1 +
1 n
n+1
单调减少有界.
由定理2.2知
:
数列1 +
1 n
n
2018/9/27
Edited by Lin Guojian
14
七大实数基本定理(它们相互等价):
1.确界定理 : 对任何非空集合E ⊂ R,如果它有上界,则必有上确界;等价地,如 果它有下界, 则必有下确界.(上确界是最小的上界, 下确界是最大的下界 ). 2.单调有界原理(即定理2.2).
3.柯西准则: 数列{an}收敛 ⇔ 对∀ε > 0, ∃N > 0,当m, n > N时, 有 am − an < ε. 4.致密性定理.
n→∞
n +1
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例
:
求
lim n→∞
1 n2 +
1
+
n2
3 +
2
+
+
2n −1 n2 + n
.
证 : n2 = 1 + 3 ++ 2n −1
n2 + n n2 + n n2 + n
n2 + n
<
1 n2 +1
+
3 n2 +
2
++
2n −1 n2 + n
5.闭区间套定理. 6.聚点定理
7.有限覆盖定理.
七大实数基本定理以不同的形式刻划了实数的连续性,即实数填满了整个 数轴,没有空隙.
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15
例 : 设数列{xn}满足 : 0 < x1 < 2, xn+1 = xn + 2, n = 1,2,.
lim
n→∞
xn
≥
0,即A
≥
0.故A
=
2.
2018/9/27
Edited by Lin Guojian
17
性质
:
设{xn
}为有界数列,
lim
n→∞
yn
=
0, 则 lim n→∞
xn
⋅
yn
=
0.
例 : 求 lim sin n ⋅ ln n
n→∞
n +1
解 : lim sin n ⋅ ln n = 0.
n2 + n
< 1 + 1 ++ 1 = n
n2 +1 n2 +1
n2 +1 n2 +1
由于 lim n = lim n = 1. n→∞ n2 + n n→∞ n2 +1
故 lni→m∞
1+ n2 +1
1 ++ n2 + 2
n
1 2+
n
=
1.
2018/9/27
Edited by Lin Guojian
nn
n
由于 lim 1 = 0,故 lim 1 ⋅ cos n = 0.
n→∞ n
n→∞ n
例 : 设xn
=
n 2(sin n)2
+ (cos
n)2 ,求 lim n→∞
xn .
解 : 由于1
≤
xn
≤
n
2,故 lim
n→∞
xn
= 1.
2018/9/27
Edited by Lin Guojian
6
定义2.1: 一个数列{an}如果满足 an ≤ an+1, n = 1,2,3,
定理2.1: (夹逼定理)假设存在正整数N0使得n ≥ N0时,数列{xn},{yn},{zn}满足
不等式 :
yn
≤
xn
≤
z
n
.如果
lim
n→∞
yn
=
lim
n→∞
zn
=
a,
那么数列{xn
}收敛且
lim
n→∞
xn
=
a.
证 :由于 lim n→∞
yn
=
a, 故对∀ε
>
0, 存在正整数N1,当n
>
N1时
有 yn − a < ε ,即a − ε < yn < a + ε.
与1
+
1 n
n+1 都收敛.
而且
lim1 +
1
n+1
=
lim 1 +
1
n
⋅ 1 +
1
=
lim1 +
1
n
⋅
lim1 +
1
n→∞ n
n→∞ n n n→∞ n n→∞ n
=
lim1
+
1
n
.
n→∞ n
2018/9/27
Edited by Lin Guojian
13
可以证明 :
证明
:
lim
n→∞
xn
存在,
并计算其极限.
证 : 首先用数学归纳法证明{xn}有界,即0 < xn < 2, n = 1,2,. 当n = 1时,由已知得 : 0 < x1 < 2,设n = k时有0 < xk < 2成立.
则当n = k +1时有0 < xk+1 = xk + 2 < 2 + 2 = 2成立.
=
A, 则 lim n→∞
xn+1
=
A.
在xn+1 = xn + 2两边取极限, 有
lim
n→∞
xn+1
=
lim
n→∞
xn + 2 ⇒ A =
A + 2 ⇒ A2 = A + 2
⇒ A2 − A − 2 = 0 ⇒ A = 2或A = −1.
由于xn > 0, n = 1,2,,由数列极限的性质2.3知
3n + 5n
=5
2018/9/27
Edited by Lin Guojian
3
例 : 设xn
=
n
3n
+ 5n
+ 9n
,求 lim n→∞
xn
证 :由于n 9n < n 3n + 5n + 9n < n 9n + 9n + 9n ⇒ 9 < n 3n + 5n + 9n < n 3⋅ 9n = n 3 ⋅ n 9n = 9 ⋅ n 3
定理 : 单调增加有上界数列必有极限.
定理 : 单调减少有下界数列必有极限.
性质 : 收敛数列{an}一定是有界数列.
2018/9/27
Edited by Lin Guojian
9
例
:
证明数列(1 +
1 n
)n
,
(1 +
1 n
) n+1 都收敛且极限相同
证 : 首先证明贝努利(Bernoulli)不等式 : 设x > −1,则(1+ x)n ≥ 1+ nx, n = 1,2,3,.
由于 lim n k = 1,故由夹逼定理知 n→∞
lim n
n→∞
a1n
+ a2n + + akn
=
A = max{a1, a2 ,ak }.
2018/9/27
Edited by Lin Guojian
5
例: 求 lim 1 ⋅ cos n n→∞ n
解 :由于 −1 ≤ cos n ≤ 1,故 − 1 ≤ 1 cos n ≤ 1 .
1 n
n+1
, n
= 1,2,为单调减少数列.
考虑 :
yn yn+1
=
1+ 1 n+1 n
1 +
1
n+2
=
n +1n+1 n n + 2 n+2
=
n +1n+1 n n + 2 n+1 ⋅ n + 2
n+1 n+1 n+1 n+1
=
(n +1)2 n(n + 2)
n+1
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8
定义 : 如果存在常数B,使得数列{an}满足 : an ≥ B, n = 1,2,, 则称{an}为有下界数列.
定义 : 如果存在常数A,使得数列{an}满足 : an ≤ A, n = 1,2,, 则称{an}为有上界数列;
定理2.2 : 单调有界数列必有极限.
< 1 + 3 ++ 2n −1 = n2
n2 +1 n2 +1
n2 +1 n2 +1
由于 lim n→∞
n2 n2 +
n
=
lim
n→∞
n2 n2 +1
= 1.
故
lim n→∞
n
1 2+
1
+
n
2
3 +
2
+
+
2n −1 n2 + n
=
1.
2018/9/27
Edited by Lin Guojian
由于 lim n 3 = 1,故由夹逼定理知 n→∞
lim
n→∞
xn
= lim n n→∞
3n + 5n
+ 9n
=9
2018/9/27
Edited by Lin Guojian
4
例 : 设ai
>
0(i
= 1,2,, k),求 lim n n→∞
a1n
+ a2n
+ + akn
解 : 设A = max{a1, a2 ,ak },则 n An ≤ n a1n + a2n + + akn ≤ n An + An + + An = n k ⋅ An ⇒ A ≤ n a1n + a2n + + akn ≤ n k ⋅ A
+ n +1)(n (n +1)3
+
2)
=
n3
+
3n2 + 3n (n +1)3
+
2
=
(n +1)3 +1 (n +1)3
> 1.
故xn+1 > xn (n = 1,2,),从而数列{xn}是单调增加数列.
2018/9/27
Edited by Lin Guojian