节: 序列:确定聚合的限定工具 | 9.2 序列:确定趋同的限定工具

章节大纲

Some of the same properties and tools that were useful in applying limits to real functions are also useful when evaluating sequences. The sum, difference, product, and quotient , as well as L’Hôpital’s Rule and the Squeeze/Sandwich Theorem all can be used to help evaluate the limits of sequences when determining their convergence. As a quick check before reading this concept, see if you can write out as many basic limit properties as possible. Do you remember L’Hôpitals’s Rule and when to use it? How about the Squeeze Theorem ?
::在对实际功能适用限制方面有用的一些相同的属性和工具在评估序列时也是有用的。总量、差异、产品和商数,以及医院规则与挤压/桑德韦奇理论都可以用来帮助评估序列的界限,以决定它们的趋同。在阅读这个概念之前,作为快速检查,请看您能否写出尽可能多的基本限制属性。您还记得医院规则及其何时使用吗? 挤压理论如何?

Tools for Determining Convergence
::确定趋同的工具

From the previous concept, you learned that determining whether a $\begin{aligned} {a_{n}} \end{aligned}$ is convergent or divergent involves evaluating $\begin{aligned} lim_{n \to \infty} a_{n} \end{aligned}$ . In this concept, we look at a few tools for evaluating some sequences for convergence, and rules that apply to convergent sequences.
::从上一个概念中,你了解到,确定 {an} 是趋同还是相异需要评估limnan。在这个概念中,我们研究一些工具来评估某些趋同序列,以及适用于趋同序列的规则。

L’Hôpital’s Rule
::医院的规则

Realistically, we cannot graph every sequence to determine if it has a finite limit and the value of that limit. Nor can we make an algebraic argument for the limit for every possible sequence. Just as there are indeterminate forms when finding limits of functions, there are indeterminate forms of sequences, such as $\begin{aligned} \frac{0}{0}, \frac{\infty}{\infty}, 0 + \infty \end{aligned}$ . To find the limit of such sequences, we can apply L’Hôpital’s rule.
::现实地说,我们无法绘制每个序列的图表来确定它是否有一定的限值和该限值的价值。我们也不能为每一种可能的序列的限值提出代数论。正如在寻找功能限制时有不确定的形式一样,也有一些不确定的序列形式,如00,,。为了找到这些序列的限值,我们可以应用L'Hital的规则。

L’Hôpital’s Rule states:
::医院规则规定:

Let functions $\begin{aligned} f \end{aligned}$ and $\begin{aligned} g \end{aligned}$ be differentiable at every number other than $\begin{aligned} c \end{aligned}$ in some interval, with $\begin{aligned} g^{'} (x) \neq 0 \end{aligned}$ if $\begin{aligned} x \neq c \end{aligned}$ . If $\begin{aligned} lim_{x \to c} f (x) = 0 \end{aligned}$ and $\begin{aligned} lim_{x \to c} g (x) = 0 \end{aligned}$ , or if $\begin{aligned} lim_{x \to c} f (x) = \pm \infty \end{aligned}$ and $\begin{aligned} lim_{x \to c} g (x) = \pm \infty \end{aligned}$ , then:
::Let f 和 g 函数在除 c 外的每个数字中都有差异, 如果 xc, 则有 g( x) =0 。如果 limx\cf( x) =0 和 limx\cg( x) =0, 或者如果 limx\cf( x) \ 和 limx\cg( x) =0, 那么 :

\begin{aligned} lim_{x \to c} \frac{f (x)}{g (x)} = lim_{x \to c} \frac{f^{' (x)}}{g^{'} (x)} \end{aligned}

x)g(x)=limx=c#(x)g(x)

as long as $\begin{aligned} lim_{x \to c} \frac{f^{' (x)}}{g^{' (x)}} \end{aligned}$ exists or is infinite.
::只要limx{cf}(x)g}(x)存在或无限。

If $\begin{aligned} f \end{aligned}$ and $\begin{aligned} g \end{aligned}$ are differentiable at every number $\begin{aligned} x \end{aligned}$ greater than some number $\begin{aligned} a \end{aligned}$ , with $\begin{aligned} g^{'} (x) \neq 0 \end{aligned}$ then:
::如果 f 和 g 在每一数字x 大于某个数字a 时是可区分的, 则 g( x) @% 0 然后 :

\begin{aligned} lim_{x \to \pm \infty} \frac{f (x)}{g (x)} = lim_{x \to \pm \infty} \frac{f^{'} (x)}{g^{'} (x)} \end{aligned}

::limxf( x) g( x) =limx\\\ f}( x) g_( x) (x)

as long as $\begin{aligned} lim_{x \to \pm \infty} \frac{f^{' (x)}}{g^{' (x)}} \end{aligned}$ exists or is infinite.
::只要limxf(x)g(x)存在或无限。

Note that L’Hôpital’s rule is also valid for .
::请注意,L ' hital的规则也适用于...。

In the previous concept, we solved %7D%7Bn%7D"> $\begin{aligned} lim_{x \to + \infty} \frac{\ln (n)}{n} \end{aligned}$ by using a graph. Let's practice using L’Hôpital’s rule instead.
::在上一个概念中,我们通过使用图表解决了lixlnnn。让我们用医院的规则来实践吧。

The sequence %7D%7Bn%7D%20%5Cright%20%5C%7D"> $\begin{aligned} {\frac{\ln (n)}{n}} \end{aligned}$ is of the indeterminate form $\begin{aligned} \frac{\infty}{\infty} \end{aligned}$ . Since the functions "> $\begin{aligned} y = \ln (n) \end{aligned}$ and $\begin{aligned} y = n \end{aligned}$ are differentiable, we can apply L’Hôpital’s rule to this problem as follows:
::{lnn} 序列为不确定的形式。由于函数 y=ln 和 y=n是不同的,我们可以对该问题适用L'Hital的规则如下:

%7D%7Bn%7D%26%3D%5Clim_%7Bn%20%5Cto%20%2B%20%5Cinfty%7D%20%5Cfrac%7B%5Cfrac%7Bd%7D%7Bdn%7D%20%5Cln%7D%7B%5Cleft(%5Cfrac%7Bd%7D%7Bdn%7D%20n%20%5Cright)%7D%20%5C%5C%0A%26%3D%5Clim_%7Bn%20%5Cto%20%2B%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Bn%7D%20%5C%5C%0A%26%3D0">

\begin{aligned} lim_{n \to + \infty} \frac{\ln (n)}{n} & = lim_{n \to + \infty} \frac{\frac{d}{d n} \ln (n)}{(\frac{d}{d n} n)} \\ = lim_{n \to + \infty} \frac{1}{n} \\ = 0 \end{aligned}

ddnn)=limn=1n=0 (ddnn)=limn=1n=0

There are some useful rules for working with sequences. Properties of function limits are also used with limits of sequences.
::使用序列有一些有用的规则,功能限制的属性也与序列限制一起使用。

Rules for Evaluating the Limit of a Sequence
::评价序列限制的规则

Let $\begin{aligned} {a_{n}} \end{aligned}$ and $\begin{aligned} {b_{n}} \end{aligned}$ be sequences such that,
::让 {an} 和 {bn} 成为这样的序列,

\begin{aligned} lim_{n \to + \infty} a_{n} = L_{1} and lim_{n \to + \infty} b_{n} = L_{2}, \end{aligned}

::利姆尼安=L1和利姆尼宾=L2

and let $\begin{aligned} c \end{aligned}$ be a constant, then:
::并保持一个常数,然后:

1. $\begin{aligned} lim_{n \to + \infty} c = c \end{aligned}$	The limit of a constant is the same constant.
2. $\begin{aligned} lim_{n \to + \infty} c \cdot a_{n} = c \cdot lim_{n \to + \infty} a_{n} = c L_{1} \end{aligned}$	The limit of a constant times a sequence is the same as the constant times the limit of the sequence.
3. $\begin{aligned} lim_{n \to + \infty} (a_{n} + b_{n}) = lim_{n \to + \infty} a_{n} + lim_{n \to + \infty} b_{n} = L_{1} + L_{2} \end{aligned}$	The limit of a sum of sequences is the same as the sum of the limits of the sequences.
4. $\begin{aligned} lim_{n \to + \infty} (a_{n} \times b_{n}) = lim_{n \to + \infty} a_{n} \times lim_{n \to + \infty} b_{n} = L_{1} L_{2} \end{aligned}$	The limit of the product of sequences is the same as the product of the limits of the sequences.
5. If $\begin{aligned} L_{2} \neq 0 \end{aligned}$ , then $\begin{aligned} lim_{n \to + \infty} (\frac{a_{n}}{b_{n}}) = \frac{lim_{n \to + \infty} a_{n}}{lim_{n \to + \infty} b_{n}} = \frac{L_{1}}{L_{2}} . \end{aligned}$	The limit of the quotient of two sequences is the same as the quotient of the limits of the sequences

To find $\begin{aligned} lim_{n \to + \infty} \frac{7 n}{9 n + 5} \end{aligned}$ we could use L’Hôpital’s rule or we could use some of the rules in the preceding theorem. Let’s use the rules in the theorem. Divide both the numerator and denominator by the highest power of $\begin{aligned} n \end{aligned}$ in the expression and using rules from the theorem, we find the limit:
::为了找到limn7n9n+5,我们可以使用医院的规则,也可以使用上一个定理中的一些规则。让我们在定理中使用规则。分子和分母除以表达中的n最高权力,使用定理中的规则,我们发现极限是:

\begin{aligned} lim_{n \to + \infty} \frac{7 n}{9 n + 5} & = lim_{n \to + \infty} \frac{7}{9 + \frac{5}{n}} & \dots Dividing both numerator and denominator by n . \\ = lim_{n \to + \infty} \frac{lim_{n \to \infty} 7}{lim_{n \to \infty} (9 + \frac{5}{n})} & \dots Applying the division rule for limits . \\ = lim_{n \to + \infty} \frac{lim_{n \to \infty} 7}{lim_{n \to \infty} 9 + lim_{n \to \infty} \frac{5}{n}} & \dots Applying the rule for the limit of a sum . \\ = lim_{n \to + \infty} \frac{7}{9 + 0} & \dots Evaluating the limits . \\ = \frac{7}{9} \end{aligned}

::==============================================================================================================================================================_==============================================================================================================================================================================================================================================================================================================================================================

Sandwich/Squeeze Theorem
::桑威奇/冷冻定理

As with limits of functions, there is a Sandwich/Squeeze Theorem for the limits of sequences that is another tool that can be used to determine whether a sequence converges .
::至于功能的限度,有一个用于序列界限的桑威奇/地震定理,这是可用来确定序列是否趋同的另一个工具。

The Sandwich/Squeeze Theorem states:
::桑威奇/冰冷定理指出:

Suppose $\begin{aligned} {a_{n}} \end{aligned}$ , $\begin{aligned} {b_{n}} \end{aligned}$ and $\begin{aligned} {c_{n}} \end{aligned}$ are sequences such that $\begin{aligned} a_{n} \leq c_{n} \leq b_{n} \end{aligned}$ for all $\begin{aligned} n \geq N \end{aligned}$ where $\begin{aligned} N \end{aligned}$ is a positive integer. If $\begin{aligned} lim a_{n} = lim b_{n} = L \end{aligned}$ , then $\begin{aligned} lim c_{n} = L \end{aligned}$ .
::假想 {an}, {bn} 和{cn} 是序列, 以 N 为正整数的 nN 表示所有 nn 。如果 liman=limbn=L, 那么 limcn=L 。

You can see how the name of the theorem makes sense from the statement. After a certain point in the sequences, the terms of a sequence $\begin{aligned} c_{n} \end{aligned}$ are sandwiched or squeezed between the terms of two convergent sequences with the same limit. Then the limit of the sequence $\begin{aligned} c_{n} \end{aligned}$ is squeezed to become the same as the limit of the two convergent sequences.
::您可以从语句中看到定理的名称是如何理解的。在顺序的某个点之后, 序列 cn 的术语被夹在或挤压在两个集合序列的术语之间, 其限制相同。然后, 序列 cn 的限度被挤压, 以便与两个集合序列的界限相同。

Let’s look at an example problem.
::让我们看看一个例子问题。

Prove $\begin{aligned} lim_{n \to + \infty} \frac{8^{n}}{n!} = 0 \end{aligned}$ .
::{\fn黑体\fs20\shad2\2aH82\3aH20\4aH33\fscx95\3cH592001\be1}证明! {\fn黑体\fs20\shad2\2aH82\3aH20\4aH33\fscx95\3cH592001\be1}

Recall that $\begin{aligned} n! \end{aligned}$ is read as “ $\begin{aligned} n \end{aligned}$ factorial” and is written as $\begin{aligned} n! = n \times (n - 1) \times (n - 2) \times \dots \times 1 \end{aligned}$ .
::提醒注意 n! 被读作“ ncistical ” , 写为 n! =nx(n-1) x(n-2)xx...x1 。

We want to apply the Sandwich theorem by squeezing the sequence $\begin{aligned} \frac{8^{n}}{n!} \end{aligned}$ between two sequences that converge to the same limit.
::我们想用桑威奇定理挤压顺序8nn! 在两个序列之间, 聚集到同一个极限。

First, we know that $\begin{aligned} 0 \leq \frac{8^{n}}{n!} \end{aligned}$ .
::首先,我们知道八零!

Now we want to find a sequence whose terms greater than or equal to the terms of the sequence $\begin{aligned} \frac{8}{7} \end{aligned}$ for some $\begin{aligned} n \end{aligned}$ .
::现在,我们想找到一个序列, 其条件大于或等于 87 序列条件的 n。

We can write
::我们可以写字

\begin{aligned} \frac{8^{n}}{n!} & = \frac{8 \times 8 \times 8 \times \dots \times 8}{n \times (n - 1) \times (n - 2) \times \dots 1} \\ = \frac{8}{n} \times \frac{8}{n - 1} \times \dots \times \frac{8}{2} \times \frac{8}{1} \\ = (\frac{8}{n}) (\frac{8}{n - 1} \times \dots \times \frac{8}{9} \times \frac{8}{8}) (\frac{8}{7} \times \frac{8}{6} \times \frac{8}{5} \times \dots \times \frac{8}{1}) \end{aligned}

::8nn!=8x8x8x8xx...x8nx(n- 1)xxx(n- 2)xx...1=8nx8n- 1xx...)x82x81=(8n)(8n- 1x...x...x89x)88(87x86x85x...x...)x81

Since each factor in the product $\begin{aligned} \frac{8}{n - 1} \times \dots \times \frac{8}{9} \times \frac{8}{8} \end{aligned}$ is less than or equal to 1, then the product $\begin{aligned} \frac{8}{n - 1} \times \dots \times \frac{8}{9} \times \frac{8}{8} \leq 1 \end{aligned}$ .
::由于产品8n-1x...x89x88中的每个系数小于或等于1,那么产品8n-1x...x89x881。

Then we make an inequality:
::然后,我们制造不平等:

\begin{aligned} (\frac{8}{n}) (\frac{8}{n - 1} \times \dots \times \frac{8}{9} \times \frac{8}{8}) (\frac{8}{7} \times \frac{8}{6} \times \frac{8}{5} \times \dots \times \frac{8}{1}) & \leq (\frac{8}{n}) (1) (\frac{8}{7} \times \frac{8}{6} \times \frac{8}{5} \times \dots \times \frac{8}{1}) \\ \leq (\frac{8}{n}) (\frac{8}{7} \times \frac{8}{6} \times \frac{8}{5} \times \dots \times \frac{8}{1}) \\ \leq (\frac{8}{n}) (\frac{8^{7}}{7!}) \end{aligned}

8n( 8n- 1x...x89x88) (87x86x85x...x81) (8)n) (1)( 87x86x85x... x81) (8)n( 8n) 87x86x85x... x81x...) (8)n( 8n) (8)x86x- 85x...) (8)x(8)x(8)1}(8)n( 8n) (8)x(8)( 877)! )

Thus, $\begin{aligned} lim_{n \to + \infty} 0 \leq lim_{n \to + \infty} \frac{8^{n}}{n!} \leq lim_{n \to + \infty} (\frac{8}{n}) (\frac{8^{7}}{7!}) \end{aligned}$ .
::如此一来,即:877!

By using the Rules Theorem, we have $\begin{aligned} lim_{n \to + \infty} 0 = 0 \end{aligned}$ and $\begin{aligned} lim_{n \to + \infty} (\frac{8}{n}) (\frac{8^{7}}{7!}) = (\frac{8^{7}}{7!}) lim_{n \to + \infty} \frac{8}{n} = (\frac{8^{7}}{7!}) \times 0 = 0 \end{aligned}$ .
::使用规则理论,我们有limn0=0和limn(8)n(8)77! (877!)=(877! ))limn8n=(877!)x0=0。

Thus, $\begin{aligned} 0 \leq lim_{n \to + \infty} \frac{8^{n}}{n!} \leq 0 \end{aligned}$ .
::因此,0月8日! 0。

By the Sandwich/Squeeze Theorem, $\begin{aligned} lim_{n \to + \infty} \frac{8^{n}}{n!} = 0 \end{aligned}$ .
::以桑威奇和冰冷的定理盟誓,

Examples
::实例

Example 1
::例1

Evaluate the following limit: $\begin{aligned} lim_{n \to \infty} \frac{n^{2}}{2^{n}} \end{aligned}$ .
::评估以下限度:limnn22n。

%3D%5Cfrac%7Bn%5E2%7D%7B2%5En%7D"> $\begin{aligned} a_{n} = f (n) = \frac{n^{2}}{2^{n}} \end{aligned}$ for $\begin{aligned} f (x) = \frac{x^{2}}{2^{x}} \end{aligned}$ .
::f( x) =x22x 的 an=f=n22n。

$\begin{aligned} lim_{x \to \infty} \frac{x^{2}}{2^{x}} \end{aligned}$ is indeterminate $\begin{aligned} \frac{\infty}{\infty} \end{aligned}$ .
::limxxx22x是不确定的。

Using L’Hôpital’s rule: $\begin{aligned} lim_{x \to \infty} \frac{x^{2}}{2^{x}} = lim_{x \to \infty} \frac{2 x}{2^{x} \cdot \ln 2} = lim_{x \to \infty} \frac{2}{2^{x} \cdot (\ln 2)^{2}} = 0 \end{aligned}$ .
::使用医院规则: limxx22x=limx2x2x2=limx22x2=0。

Therefore, $\begin{aligned} lim_{n \to \infty} \frac{n^{2}}{2^{n}} = 0 \end{aligned}$ .
::因此, limnn22n=0。

Review
::回顾

Tell if each sequence is convergent, is divergent, or has no limit. If the sequence is convergent, find its limit.
::告诉每个序列是集合的, 是不同的, 还是没有限制的。如果序列是集合的, 请找到它的极限。

$\begin{aligned} lim_{n \to \infty} \sqrt[8 n]{3 n} \end{aligned}$
::立方公尺三nn

$\begin{aligned} lim_{n \to \infty} \frac{3 n^{2} - 5}{n^{2} - 5 n + 2} \end{aligned}$
::立米3n2-5n2-5n2-5n5+2

$\begin{aligned} lim_{n \to \infty} \frac{n}{e^{n}} \end{aligned}$
::里姆诺

%7D%7Bn%5E2%7D"> $\begin{aligned} lim_{n \to \infty} \frac{\ln (n)}{n^{2}} \end{aligned}$
:n)n2

$\begin{aligned} lim_{n \to \infty} {(1 - \frac{b}{n})}^{a n} \end{aligned}$
::立米(1-bn)an

$\begin{aligned} lim_{n \to \infty} \frac{n}{n^{2} + 1} \end{aligned}$
::limnn2+1

$\begin{aligned} lim_{n \to \infty} \frac{\cos n}{n} \end{aligned}$
::来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

$\begin{aligned} lim_{n \to \infty} \frac{\sin n}{n} \end{aligned}$
::来来来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

$\begin{aligned} lim_{n \to \infty} \frac{7}{n!} \end{aligned}$
::来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来!

$\begin{aligned} lim_{n \to \infty} (\frac{n^{2}}{n + 1} - \frac{1}{n + 1}) \end{aligned}$
::limn(n2n+1-1n+1)

$\begin{aligned} lim_{n \to \infty} \frac{\ln (n + 7)}{n^{\frac{1}{n}}} \end{aligned}$
:n+7)n1nn (n+7)nn

$\begin{aligned} lim_{n \to \infty} \frac{n^{3} + n^{2} - n + 1}{n^{2} + 1} \end{aligned}$
::立方公尺N3+n2-n+1n2+1

$\begin{aligned} lim_{n \to \infty} \frac{\sin^{2} n}{3^{n}} \end{aligned}$
::来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

$\begin{aligned} lim_{n \to \infty} (3 - \frac{1}{3^{n}}) (2 + \frac{1}{2^{n}}) \end{aligned}$
:3-13n)(2+12n)

$\begin{aligned} lim_{n \to + \infty} (\frac{11}{n} - \frac{8}{n^{2}}) \end{aligned}$
::立米(11n-8n2)

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

Tools for Determining Convergence ::确定趋同的工具

L’Hôpital’s Rule ::医院的规则

Rules for Evaluating the Limit of a Sequence ::评价序列限制的规则

Sandwich/Squeeze Theorem ::桑威奇/冷冻定理

Examples ::实例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)