Course: 9.1 顺序、限制和趋同

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 顺序、限制和趋同

Collapse Expand
顺序、限制和趋同
The alphabet, the names in a phone book, the numbered instructions of a model airplane kit, and the schedule in the local television guide are examples of sequences. These examples are all sets of ordered items. In mathematics, a can be a list of numbers. The list of numbers can be finite or infinite. The list can be generated based on a known mathematical rule, or just randomly generated. There are many well known infinite number sequences. See if you recognize and can name this famous one: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . . . Can you state the mathematical rule that generates it?
::字母表、电话簿中的名称、模拟飞机工具包的编号指示, 以及本地电视指南中的时间表是序列的示例。这些示例都是有顺序的项目。在数学中, 这些示例都是有顺序的项目组。在数学中, 数字列表可以是数字列表。数字列表可以是有限或无限的。列表可以根据已知的数学规则生成, 也可以是随机生成。有很多已知的无限数序列。请看您是否识别并可以命名这个著名的序列 : 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 。您能否说明生成该列表的数学规则 ?

Sequences
::序列

The alphabet, the names in a phone book, the numbered instructions of a model airplane kit, and the schedule in the local television guide are examples of sequences. These examples are all sets of ordered items. In mathematics, a sequence can be a list of numbers. There are finite sequences, such as 2, 4, 6, 8. These sequences end. There are infinite sequences, such as 3, 5, 7, 9, . . . , which do not end but continue on as indicated by the three dots. In this and subsequent concepts the word sequence will refer to an infinite sequence.
::字母表、电话簿中的名称、模拟飞机装具的编号指示, 以及本地电视指南中的时间表是序列的示例。这些示例是所有组的定序项。在数学中, 序列可以是数字列表。有一定的序列, 如 2 、 4 、 6 、 8 。这些序列的结尾。有无限的序列, 如 3 、 5 、 7 、 9 . . , 它们不会结束, 但会像三个点所显示的那样继续下去。在此以及随后的概念中, 单词序列将指一个无限的序列。

A sequence, denoted by $\begin{aligned} {a_{n}} \end{aligned}$ or by $\begin{aligned} a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{n}, \dots \end{aligned}$ , is a function whose domain is the set of positive integers $\begin{aligned} n \end{aligned}$ , and whose range consists of the terms of the sequence $\begin{aligned} a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{n}, \dots \end{aligned}$ .
::由 {an} 或 a1,a2,a3,a4,a4 表示的序列, 是一个函数, 其域是正数整数 n 的一组, 其范围包含序列 a1,a2,a3,a4,a4的参数。

Each subscript of 1, 2, 3, . . . on the terms $\begin{aligned} a_{1}, a_{2}, a_{3}, a_{4}, \dots \end{aligned}$ is an index that refers to the place of the term in the sequence. The subscripts are called the indices of the terms. We assume that $\begin{aligned} n = 1, 2, 3, \dots, \end{aligned}$ unless otherwise noted.
::每一下标 1, 2, 3, . . 在 a1, a2, a3, a4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Instead of listing the elements of a sequence, a sequence can often be defined by a rule, or formula, in terms of the indices.
::与其列出序列要素,不如按指数用规则或公式来界定序列。

Each term in a sequence is defined by its place of order in the list. Consider the sequence 3, 5, 7, 9, . . . . The first term is 3 because it belongs to place 1 of the sequence. The second term is 5 because it belongs to the second place of the sequence. Likewise, the third term is 7 because it is in the third place.
::顺序中每个术语的顺序由列表中的顺序位置决定。考虑顺序 3 、 5 、 7 、 9 . . . 。第一个术语为 3 , 因为它属于序列的第 1 位。第二个术语为 5 , 因为它属于序列的第二位。同样, 第三术语为 7 , 因为它位于第三位。

For practice, let's g enerate the terms of the sequence using the rule for $\begin{aligned} a_{n} = \frac{1}{n} \end{aligned}$ .
::为了实践,让我们用 an=1n 的规则来生成序列的条件。

We can generate the terms for the rule $\begin{aligned} a_{n} = \frac{1}{n} \end{aligned}$ as follows:
::我们可以产生以下规则a=1n的条件:

$\begin{aligned} n & 1 & 2 & 3 & 4 & \dots \\ a_{n} = \frac{1}{n} & \frac{1}{1} = 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \dots \end{aligned}$

::n1234... an=1n11=11121314...

It is often more difficult to generate a rule than to write terms. Let's generate the rule that matches the sequence: $\begin{aligned} \frac{1}{2}, - \frac{2}{3}, \frac{3}{4}, - \frac{4}{5}, \dots \end{aligned}$ .
::产生规则往往比写术语更困难。让我们产生与顺序相符的规则:12,-23,34,-45,......。

$\begin{aligned} n & 1 & 2 & 3 & 4 \\ a_{n} & \frac{1}{2} & - \frac{2}{3} & \frac{3}{4} & - \frac{4}{5} \end{aligned}$

::n1234an12-3334-45

Look at each term in terms of its index. The numerator of each term matches the index. The denominator is one more than the index. So far, we can write the formula $\begin{aligned} a_{n} \end{aligned}$ as $\begin{aligned} \frac{n}{n + 1} \end{aligned}$ .
::以索引的形式查看每个术语。每个术语的分子与索引相匹配。分母比索引多一个。到目前为止, 我们可以将公式写成 nn+1 。

However, we are not done. Notice that each even-indexed term has a negative sign. This means that all of terms of the sequence have a power of -1. The powers of -1 alternate between odd and even. Usually, alternating powers of -1 can be denote by $\begin{aligned} (- 1)^{n} \end{aligned}$ or $\begin{aligned} (- 1)^{n + 1} \end{aligned}$ .
::然而,我们没有完成。请注意, 每个偶数索引术语都有负信号, 这意味着序列中的所有术语都具有 -1. 奇数和偶数之间 -1 的交替权力。通常, 1- 1 的交替权力可以由 (-1)n 或 (-1)n+1 表示。

Since the terms are negative for even indices, we use $\begin{aligned} (- 1)^{n + 1} \end{aligned}$ . Thus, the rule for the sequence is $\begin{aligned} a_{n} = \frac{(- 1)^{n + 1} n}{n + 1} \end{aligned}$ . You can check the rule by finding the first few terms of the sequence.
::由于单指数的术语为负值,所以我们使用(-1)n+1。因此,序列的规则是 an=(-1)n+1nn+1。您可以通过找到序列的前几个术语来检查该规则。

Limit of a Sequence
::序列限制

We are interested in the behavior of the sequence as the value of $\begin{aligned} n \end{aligned}$ gets very large. Many times a sequence will get closer to a certain number, or limit, as $\begin{aligned} n \end{aligned}$ gets large. Finding the limit of a sequence is very similar to finding the limit of a function. Let’s look at some graphs of sequences.
::当 n 值变得非常大时,我们对该序列的行为很感兴趣。许多次序列会越来越接近某个数字,或者限制到 n 值的大小。找到一个序列的界限与找到函数的界限非常相似。让我们看看一些序列的图表。

If we were to try to find the limit of the sequence $\begin{aligned} {\frac{1}{2 n + 1}} \end{aligned}$ as $\begin{aligned} n \end{aligned}$ goes to infinity, a graph helps to visualize what is happening.
::如果我们试图找到 n 到无穷的序列 {12n+1} 的极限, 图表有助于直观地描述正在发生的事情。

We can graph the corresponding function $\begin{aligned} y = \frac{1}{2 n + 1} \end{aligned}$ for $\begin{aligned} n = 1, 2, 3, \dots \end{aligned}$ . The graph is similar to the continuous function $\begin{aligned} y = \frac{1}{2 x + 1} \end{aligned}$ for the domain of $\begin{aligned} x \geq 1 \end{aligned}$ .
::我们可以为 n= 1, 2, 3,...... 绘制相应的函数 y= 12n+1 。该图与 x= 1 域的连续函数 y= 12x+1 相似。

To determine the limit, we look at the trend or behavior of the graph of sequence as $\begin{aligned} n \end{aligned}$ gets larger or travels out to positive infinity. This means we look at the points of sequence that correspond to the far right end of the horizontal axis in the figure on the right. We see that the points of the sequence are getting closer to the horizontal axis, $\begin{aligned} y = 0 \end{aligned}$ . Thus, the limit of the sequence $\begin{aligned} {\frac{1}{2 n + 1}} \end{aligned}$ is 0 as $\begin{aligned} n \end{aligned}$ tends to infinity. We write: $\begin{aligned} lim_{n \to + \infty} \frac{1}{2 n + 1} = 0 \end{aligned}$ .
::为确定限制, 我们以 n 变大或向正无穷来查看序列图的趋势或行为。这意味着我们查看与右边图中水平轴极右端相对应的序列点。我们看到序列点正在接近水平轴, y=0 。因此, 序列 { 12n+1} 的极限是 0 n , 与无穷一样。我们写 : limn 12n+1=0 。

Here is the precise definition of the limit of a sequence:
::以下是对序列界限的精确定义:

The limit of a sequence $\begin{aligned} a_{n} \end{aligned}$ is the number $\begin{aligned} L \end{aligned}$ such that if for each $\begin{aligned} ε > 0 \end{aligned}$ , there exists an integer $\begin{aligned} N \end{aligned}$ such that $\begin{aligned} | a_{n} - L | < ε \end{aligned}$ for all $\begin{aligned} n > N \end{aligned}$ .
::序列的限值是数字L,因此,如果每 +++0,则存在整数N,即所有 n>N 的 an-L = an-L=N。

$\begin{aligned} | a_{n} - L | < ε \end{aligned}$ means the values of $\begin{aligned} a_{n} \end{aligned}$ such that $\begin{aligned} L - ε < a_{n} < L + ε \end{aligned}$ .
::an-L意指Lan<L的价值观。

Each sequence’s limit falls under only one of the four possible cases:
::每个序列的限值仅属于四种可能情况中的一种:

A limit exists and the limit is $\begin{aligned} L \end{aligned}$ : $\begin{aligned} lim_{n \to + \infty} a_{n} = L \end{aligned}$ .
::限值存在,限值为L:limnan=L。

There is no limit: $\begin{aligned} lim_{n \to + \infty} a_{n} \end{aligned}$ does not exist .
::不存在限制: limnan不存在。

The limit grows without bound in the positive direction: $\begin{aligned} lim_{n \to + \infty} a_{n} = + \infty \end{aligned}$ .
::限制在不拘泥于积极方向的情况下增长: limnan。

The limit grows without bound in the negative direction: $\begin{aligned} lim_{n \to + \infty} a_{n} = - \infty \end{aligned}$ .
::限制在不受负面约束的情况下增长: limnan。

Figure below shows the graph of the sequence %7D%7Bn%7D%20%5Cright%20%5C%7D"> $\begin{aligned} {\frac{l_{n} (n)}{n}} \end{aligned}$ .
::下图显示序列 {lnn} 的图表。

Notice that from $\begin{aligned} N \end{aligned}$ on, the terms of $\begin{aligned} \frac{l_{n} n}{n} \end{aligned}$ are between $\begin{aligned} L - ε \end{aligned}$ and $\begin{aligned} L + ε \end{aligned}$ . In other words, for this value of $\begin{aligned} ε \end{aligned}$ , there is a value $\begin{aligned} N \end{aligned}$ such that all terms of $\begin{aligned} a_{n} \end{aligned}$ are in the interval from $\begin{aligned} L - ε \end{aligned}$ and $\begin{aligned} L + ε \end{aligned}$ . Thus, %7D%7Bn%7D%3D0"> $\begin{aligned} lim_{n \to + \infty} \frac{l_{n} (n)}{n} = 0 \end{aligned}$ .
::请注意,从N到N, 内恩的条件介于L和L之间。换句话说, 对于这个值, 有一个值是N, 所以所有条件都介于L和L之间的间距。因此, limnlnn=0 。

Not every sequence has a limit.
::并不是每个序列都有极限

Consider the sequence $\begin{aligned} {n + 1} \end{aligned}$ in Figure below.
::考虑下图中的顺序 {n+1} 。

As $\begin{aligned} n \end{aligned}$ gets larger and goes to infinity, the terms of $\begin{aligned} a_{n} = n + 1 \end{aligned}$ become larger and larger. The sequence $\begin{aligned} {n + 1} \end{aligned}$ does not have a limit. We write $\begin{aligned} lim_{n \to + \infty} (n + 1) = + \infty \end{aligned}$ .
::n 随着 n 的扩大和无穷, an=n+1 的条件会变大。序列 {n+1} 没有限制。我们写 limn(n+1)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Convergence and Divergence
::趋同和相异

We say that a sequence $\begin{aligned} {a_{n}} \end{aligned}$ converges if the sequence has a finite limit $\begin{aligned} L \end{aligned}$ . The sequence then has convergence ; it converges to the limit $\begin{aligned} L \end{aligned}$ , and we describe the sequence as convergent.
::我们说,如果序列有限制的L.,则序列{an}会趋同。然后,序列会趋同;它会同到限制的L,我们把序列描述为趋同。

If a sequence is convergent, then its limit is unique.
::如果一个序列是趋同的,那么它的极限是独特的。

On the other hand, if the limit of a sequence $\begin{aligned} {a_{n}} \end{aligned}$ grows without bound in either the positive or negative direction the sequence is said to diverge . The sequence has divergence and we describe the sequence as divergent. Keep in mind that being divergent is not the same as not having a limit.
::另一方面,如果序列 {an} 的极限在不受约束的情况下以正或负方向增长,则该序列据说是不同的。序列有差异,我们将序列描述为不同。要记住,分歧并不等于没有限制。

Using the information above, let's determine whether the sequences converge :
::使用上述信息,让我们确定顺序是否趋同:

1. %20%5C%7D"> $\begin{aligned} {a_{n} = \ln (n)} \end{aligned}$
::1. {an=ln{}

2. $\begin{aligned} {a_{n} = 4 - \frac{8}{n}} \end{aligned}$
::2. {an=4-8n}

3. $\begin{aligned} 1, - 1, 1, - 1, 1, - 1, \dots \end{aligned}$ .

The sequence %20%5C%7D"> $\begin{aligned} {\ln (n)} \end{aligned}$ grows without bound as $\begin{aligned} n \end{aligned}$ approaches infinity. Note that the related function $\begin{aligned} y = \ln (x) \end{aligned}$ grows without bound. The sequence is divergent because it does not have a finite limit. We write %3D%2B%20%5Cinfty"> $\begin{aligned} lim_{n \to + \infty} \ln (n) = + \infty \end{aligned}$ .
::序列 {ln} 不受 n 无限接近的约束而成长。请注意, 相关的函数 y=ln(x) 不受约束而成长。序列是不同的, 因为它没有限定限制。我们写着 limn ln\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

The sequence $\begin{aligned} {a_{n} = 4 - \frac{8}{n}} \end{aligned}$ converges to the limit $\begin{aligned} L = 4 \end{aligned}$ and hence is convergent. If you graph the function $\begin{aligned} y = 4 - \frac{8}{n} \end{aligned}$ for $\begin{aligned} n = 1, 2, 3, \dots \end{aligned}$ , you will see that the graph approaches 4 as $\begin{aligned} n \end{aligned}$ gets larger. Algebraically, as $\begin{aligned} n \end{aligned}$ goes to infinity, the term $\begin{aligned} - \frac{8}{n} \end{aligned}$ gets smaller and tends to 0 while 4 stays constant. We write $\begin{aligned} lim_{n \to + \infty} (4 - \frac{8}{n}) = 4 \end{aligned}$ .
::{an=4-8n} 序列会合到 L=4 的限值, 因此会趋同。如果您用 n= 1, 2, 3, ... 来绘制 y=4-8n 函数, 则会看到 y=4- 8n 的 y= 1, 2, 3, 3, ... 的 y= 4- 8n 函数会变大。代数上, 如 n 到无穷, 则该词会变小, 至 8n 的值会变小, 并会变小为 0, 而 4 则保持不变。我们写 limn\\\ (4- 8n) = 4 4 。

The sequence $\begin{aligned} 1, - 1, 1, - 1, 1, - 1, \dots \end{aligned}$ . oscillates, or goes back and forth, between the values 1 and -1. The sequence does not get closer to 1 or -1 as $\begin{aligned} n \end{aligned}$ gets larger. We say that the sequence does not have a limit, or $\begin{aligned} lim_{n \to + \infty} S_{n} \end{aligned}$ does not exist.
::序列 1 - 1, 1, 1, 1- 1, .... 振动, 或者回转, 在值 1 和 - 1 之间。序列不会随着 n 变大而接近 1 或 - 1 。我们说序列没有限制, 或者 limnSn 不存在。

Examples
::实例

Example 1
::例1

Earlier, you were asked to state the mathematical rule that generates the following sequence:
::先前, 您被要求描述生成以下序列的数学规则 :

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 . . . .

Were you able to recognize this famous sequence that has been found in nature to describe the leaf arrangement in plants, the pattern of the florets of a flower, the bracts of a pinecone, or the scales of a pineapple. These numbers of the Fibonacci sequence are applicable to the growth of many living things.
::如果你能够认出这个著名的序列, 这个在自然界中发现的序列来描述植物的叶子结构,花叶的形态, 松树的形态, 菠萝的形态, 或菠萝的大小。 Fibonacci 序列的这些数字适用于许多生物的生长。

By definition, the first two numbers in the Fibonacci sequence are 0 and 1, and each subsequent number is the sum of the previous two. In mathematical terms, the sequence $\begin{aligned} F_{n} \end{aligned}$ of Fibonacci numbers is defined by the recurrence relation: $\begin{aligned} F_{n} = F_{n - 1} + F_{n - 2} \end{aligned}$ , with $\begin{aligned} F_{0} = 0, F_{1} = 1 \end{aligned}$ .
::根据定义,Fibonacci序列中前两个数字是0和1, 其后的每个数字是前两个数字的总和。从数学角度来说, Fibonacci 数字的Fn 序列由复发关系定义: Fn=Fn-1+Fn-2, F0=0, F1=1。

The Fibonacci sequence is named after . His 1202 book introduced the sequence to Western European mathematics, although the sequence had been described earlier in . By modern convention, the sequence begins either with $\begin{aligned} F_{0} = 0 \end{aligned}$ or with $\begin{aligned} F_{1} = 1 \end{aligned}$ .
::Fibonacci 序列是按此命名的。他的 1202 个书本将序列引入西欧数学, 尽管此序列在前面已有描述。根据现代惯例, 该序列要么以 F0=0or 和 F1=1 开始。

Example 2
::例2

Use the $\begin{aligned} n \end{aligned}$ th term rule to generate the indicated terms in each sequence:
::使用 nth 术语规则在每个序列中生成指定的术语 :

$\begin{aligned} a_{n} = 2^{n} - 1 \end{aligned}$ ; terms 1-3 and the 10 ^th term.
::a=2n-1;任期1-3和第10届。

For $\begin{aligned} a_{n} = 2^{n} - 1 \end{aligned}$ , the first three terms are: $\begin{aligned} 2^{1} - 1 = 1, 2^{2} - 1 = 3, 2^{3} - 1 = 7 \end{aligned}$ ; $\begin{aligned} a_{10} = 2^{10} - 1 = 1023 \end{aligned}$ .
::Foran=2n-1,前三个学期为:21-1=1,22-1=3,23-1=7;a10=210-1=1023。

Example 3
::例3

Write the $\begin{aligned} n \end{aligned}$ th term rule for the sequence 1, 7, 25, 79, . . . .
::第1、7、25、79、...

Examination of the sequence 1, 7, 25, 79, . . . . shows that the differences between successive terms are 6, 18, and 54. These can be written as $\begin{aligned} 6 \cdot 3^{0}, 6 \cdot 3^{1}, 6 \cdot 3^{2} \end{aligned}$ or $\begin{aligned} 2 \cdot 3^{1}, 2 \cdot 3^{2}, 2 \cdot 3^{3} \end{aligned}$ . A recursive rule that states $\begin{aligned} a_{n} - a_{n - 1} = 2 \cdot 3^{n - 1} \end{aligned}$ would work as long as for $\begin{aligned} n = 1 \end{aligned}$ , $\begin{aligned} a_{n - 1} = - 1 \end{aligned}$ . The rule would be $\begin{aligned} a_{n} = a_{n - 1} + 2 \cdot 3^{n - 1} \end{aligned}$ with $\begin{aligned} a_{0} = - 1 \end{aligned}$ .
::对顺序1、7、25、79、.的检查表明,连续任期之间的差异为6、18和54,可以写成6-30、6-31、6-32或2-31、2-32、2-23-33. 一条累回溯性规则,即只要福恩=1、an-1-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

Example 4
::例4

Determine if the sequence $\begin{aligned} {\frac{n^{2} - 2 n + 3}{3 n^{2} + 6 n - 4}} \end{aligned}$ is convergent, is divergent, or has no limit. If the sequence is convergent, find its limit.
::确定 {n2- 2n+33n2+6n-4} 序列是否趋同、有差异或没有限制。如果该序列是趋同,请找到其极限。

$\begin{aligned} lim_{n \to \infty} \frac{n^{2} - 2 n + 3}{3 n^{2} + 6 n - 4} = lim_{n \to \infty} \frac{n^{2} (1 - \frac{2}{n} + \frac{3}{n^{2}})}{n^{2} (3 + \frac{6}{n} - \frac{4}{n^{2}})} = \frac{1}{3} \end{aligned}$ . The sequence is convergent to the value $\begin{aligned} \frac{1}{3} \end{aligned}$ .
::n2 - 2n+33n2+6n4=limn@n2(1-2n+3n2)n2(3+6n-4n2)=13。序列与值一致13。

Review
::回顾

For #1-3, use the given $\begin{aligned} n \end{aligned}$ th term rule to generate the indicated terms in the sequence.
::对于 # 1-3, 使用给定的 nth 术语规则在序列中生成指定的术语。

$\begin{aligned} 2 n + 7 \end{aligned}$ , for terms 1-5 and the 10 ^th term.
::2n+7,任期1-5和第10届。

$\begin{aligned} {(\frac{1}{2})}^{n} \end{aligned}$ , for terms 1-3 and the 8 ^th term.
:12) 任期1至3和第8年。

$\begin{aligned} \frac{n (n + 1)}{2} \end{aligned}$ , for terms 1-4 and the 20 ^th term.
::n(n+112),任期1-4和20年。

For #4-6, write the $\begin{aligned} n \end{aligned}$ th term rule for the sequences.
::对于#4-6, 写下Nth术语规则用于序列。

-2, 2, -2, 2, . . .

6, 14, 24, 36, . . .

2, 5, 9, 14, . . .

For #7-14, tell if each sequence is convergent, is divergent, or has no limit. If the sequence is convergent, find its limit.
::对于 # 7-14, 请说明每个序列是否趋同、不同或没有限制。如果序列是趋同, 请找到其极限。

$\begin{aligned} {\frac{4}{n} + \frac{3}{n^{2}}} \end{aligned}$
::{4n+3n2}

$\begin{aligned} {6 - \frac{7}{\sqrt{n}}} \end{aligned}$
::{6-7n}

-5, 5, -5, 5, -5, 5, . . .

$\begin{aligned} {\frac{4 n^{6} - 7}{3 n}} \end{aligned}$
::{4n6-73n}

$\begin{aligned} {\frac{(- 1)^{n}}{5 n^{2}}} \end{aligned}$
::{(- 1) n5n2}

$\begin{aligned} {(- 1)^{n} n} \end{aligned}$
::{( - 1) nn}

$\begin{aligned} {(- 1)^{n} \frac{3 n^{4} - 2}{2 n^{4} + 6 n^{2} - 4 n}} \end{aligned}$
::{(-1)n3n4-22n4+6n2-4n}

$\begin{aligned} {\frac{6 n^{2}}{e^{n}}} \end{aligned}$
::{6n2en} {6n2en} {6n2en}

Let $\begin{aligned} {a_{n}} \end{aligned}$ be a sequence such that $\begin{aligned} lim_{n \to + \infty} | a_{n} | = 0 \end{aligned}$ . Show that $\begin{aligned} lim_{n \to + \infty} a_{n} = 0 \end{aligned}$ . ( $\begin{aligned} | a_{n} | \end{aligned}$ is the absolute value of $\begin{aligned} a_{n} \end{aligned}$ .)
::让{an} 成为这样的序列, 使 limnan_0. 显示 limnan=0. (an是 a 的绝对值。)

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。

Section outline

General

顺序、限制和趋同

Sequences ::序列

Limit of a Sequence ::序列限制

Convergence and Divergence ::趋同和相异

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Sequences
::序列

Limit of a Sequence
::序列限制

Convergence and Divergence
::趋同和相异

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)