Course: 9.7 正值序列:比较试验

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阳性术语序列:比较测试

The comparison test that is considered in this concept is based on the ideas that (1) if a positive term series is always greater, term by term, than another infinite series that diverges , than the positive term series must also diverge , and (2) if a positive term series is always smaller, term by term, than another infinite series that converges , than the positive term series must also converge . If $\begin{aligned} △_{n} \end{aligned}$ represents how much greater or smaller a term $\begin{aligned} a_{n} \end{aligned}$ of a series is compared to a term $\begin{aligned} b_{n} \end{aligned}$ of a series of known convergence or divergence, can you formulate expressions that show (1) and (2)?
::这一概念中考虑的比较检验标准基于以下想法1) 如果一个正用语系列总是比另一个千差万别无穷的系列大,按术语逐年计算,则肯定用语系列也必定不同;(2) 如果一个正用语系列总比另一个相交的无限系列小,按术语逐年计算,则肯定用语系列也比正用语系列要合并。如果 n 表示一个系列的一个术语比一系列已知趋同或分歧的一亿个术语大多少小,您能否提出显示(1)和(2)的表达方式?

Comparison Tests
::比较测试

Series with only nonnegative terms, i.e., terms that are either positive or zero, are often called positive term series, and are described as $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ with $\begin{aligned} u_{k} \geq 0 \end{aligned}$ for every $\begin{aligned} k \end{aligned}$ . Several of the types of series identified in the previous concepts are, or can be, nonnegative (or positive) term series as shown below:
::仅使用非负值术语的系列,即正值或零值的术语,通常称为正值术语序列,被描述为+k=1uk,每k.k.k.k.k.k.k.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx/或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,或正值,通常为正值,通常为正值序列,通常称为正值序列,通常为正值序列,通常称正值,且称正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且称为正值,且为正值,且称正值,且称正值,且称正值,且称正值,且称正值,且为正值为正值,且为正值,且为正值,且为正值,且为正值,且为正值,且为正值为正值为正值,且为正,且为正,且为正

**Common Series Types can be (are) Positive Term Series**
Series ::系列丛书系列 Type ::类型类型类型	Sigma Notation ::Sigma 符号	Converges if ::组合	Diverges if ::如果	Positive ::阳 Term Series if ::系列(如果)
Arithmetic ::测量学	$\begin{aligned} S = \sum_{n = 1}^{\infty} [t_{0} + d (n - 1)] \end{aligned}$	Never ::从未	Always ::总是	$\begin{aligned} t_{0}, d \geq 0 \end{aligned}$
Geometric ::几何	$\begin{aligned} S = \sum_{n = 1}^{\infty} a r^{n - 1} \end{aligned}$	$\begin{aligned} \| r \| < 1 \end{aligned}$ ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ with ::与 $\begin{aligned} S = \frac{a}{1 - r} \end{aligned}$ ::S=a1-r	$\begin{aligned} \| r \| \geq 1 \end{aligned}$ ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\	$\begin{aligned} a, r \geq 0 \end{aligned}$
Harmonic ::调音	$\begin{aligned} S = \sum_{n = 1}^{\infty} \frac{1}{n} \end{aligned}$	Never ::从未	Always ::总是	Always ::总是
$\begin{aligned} p \end{aligned}$ -Series ::p- 系列	$\begin{aligned} S = \sum_{n = 1}^{\infty} \frac{1}{n^{p}}, p > 0 \end{aligned}$ ::Sn=11np,p>0	$\begin{aligned} p > 1 \end{aligned}$	$\begin{aligned} 0 < p \leq 1 \end{aligned}$ ::0<p1	Always ::总是

So far we have looked at the following tests for the convergence/divergence of infinite series:
::迄今为止,我们研究了下列无限系列趋同/共振的测试:

Convergence/Divergence Test ::趋同/活力试验	Applicable Series ::适用系列
Limit of $\begin{aligned} n \end{aligned}$ th partial sum ::Nn部分金额限额	All ::全部
$\begin{aligned} n \end{aligned}$ th-term test for divergence ::nth 差异期度测试	All ::全部
	Positive term ::阳性

The following tests are specifically made for evaluating positive term series:
::专门为评价正值系列进行了下列测试:

The Integral Test
::综合综合测试

Comparison Tests (the Basic, The Simplified Limit Comparison Test)
::比较测试(基础,简化的限制性比较测试)

Ratio and Root Tests
::比率和根测试

This concept will focus on several comparison tests, i.e. tests that compare one infinite series of unknown convergence with another series of known convergence. The comparison can be term by term, or via the ratio of terms.
::这一概念将侧重于若干比较测试,即将一系列未知趋同与另一系列已知趋同进行比较的测试。比较可以按术语或按术语比进行。

The (Direct) Comparison Test
:直接)比较测试

The name of the test tells us that we will compare one series to another to determine convergence or divergence.
::试验的名称告诉我们,我们将比较一个系列和另一个系列,以确定趋同或分歧。

The (Direct) Comparison Test is as follows:
:直接)比较测试如下:

Suppose $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ are series with non-negative terms, then:
::假设“k”=1uk和“k”=1vk为非负值术语序列,然后:

If $\begin{aligned} u_{k} \leq v_{k} \end{aligned}$ for every positive integer $\begin{aligned} k \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ converges, then $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ converges.
::如果每个正整数 k 和 k= 1\\\ vk 相交, 则 k= 1\\\ uk 相交。

If $\begin{aligned} u_{k} \geq v_{k} \end{aligned}$ for every positive integer $\begin{aligned} k \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ diverges, then $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ diverges.
::如果对每个正整数 k 和 k= 1\\\ vk 差数 ukvk, 那么 k= 1\\ uk 差数。

In order to use this test, we must check the relationship between $\begin{aligned} u_{k} \end{aligned}$ and $\begin{aligned} v_{k} \end{aligned}$ for each index $\begin{aligned} k \end{aligned}$ . This is the comparison part of the test. If the series with the greater-valued terms converges, than the series with the lesser-valued terms converges. If the lesser-valued series diverges, then the greater-valued series will diverge.
::为了使用此测试, 我们必须检查每个指数 k 的 uk 和 vk 之间的关系。这是测试的比较部分。如果与价值更高的条件相匹配的序列比与价值较低的条件相融合的序列更为一致, 那么价值较低的序列就会有所不同。

Let's determine whether $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3} + 3} \end{aligned}$ converges or diverges.
::让我们来决定 k=1\\\\\\\\\\\1k3+3 是否趋同或不同。

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3} + 3} \end{aligned}$ looks similar to $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3}} \end{aligned}$ , so we will try to apply the Comparison Test.
::k=111k3+3 看上去类似于 k=11k3, 所以我们将尝试应用比较测试。

First compare each term of both series: for each $\begin{aligned} k \end{aligned}$ , $\begin{aligned} \frac{1}{k^{3} + 3} < \frac{1}{k^{3}} \end{aligned}$ so $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3} + 3} < \sum_{k = 1}^{\infty} \frac{1}{k^{3}} \end{aligned}$ .
::首先比较两个序列的每个术语: 对于每个 k, 1k3+3 < 1k3 so k=1\\\\\\\1\1k3\3\3\3\3\3\1\1\1k3\1k3\1\1k3, 每个 k, 1k3+3+3 < 1k3 so k=1\1\1k3\1k3\1\1k3 。

Next, we know that $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3}} \end{aligned}$ is a $\begin{aligned} p \end{aligned}$ -series that converges because $\begin{aligned} p > 1 \end{aligned}$ .
::接下来,我们知道"k=1"1"1k3 是一个P系列, 因为p>1而聚集在一起。

Therefore, by the Comparison Test, $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{k^{3} + 3} \end{aligned}$ also converges.
::因此,根据比较测试,k=11k3+3也趋于一致。

The Limit Comparison Test
::限制性比较测试

The Limit Comparison Test is easier to use than the Comparison Test for determining the convergence of series non-negative terms.
::在确定非负数系列术语的趋同方面,使用限制比较试验比比较试验容易。

The Limit Comparison Test is as follows:
::限制性比较测试如下:

Suppose $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ are series with $\begin{aligned} u_{k} > 0 \end{aligned}$ and $\begin{aligned} v_{k} > 0 \end{aligned}$ for all $\begin{aligned} k \end{aligned}$ ,
::假设“k”=1uk 和“k”=1vk 与 uk>0 和 vk>0 的序列为所有 k,

then:
::然后:

If $\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} = L \end{aligned}$ , where $\begin{aligned} 0 < L < \infty \end{aligned}$ , then $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ both converge or both diverge.
::如果 limkáukvk=L, 位于 0<L, 那么k=1uk 和k=1vk 两者都趋同, 或者两者都有分歧。

If $\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} = 0 \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ converges, then $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ converges.
::如果立方ukvk=0和k=1vk交汇,则立方k=1uk交汇。

If $\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} = + \infty \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ diverges, then $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ diverges.
::如果立方公尺和立方公尺差1公尺差,那么立方公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差差差差1公尺差1公尺差1公尺差差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公尺差1公差1公差1公差1公差1公差1公差1公差1公差1公差差1公差1公差1公差1公差1公差差差1公差1公差差差1公差1公差1公差1公差1公差1公差1公差1公差1公差1公差1公差1公差1公差1公差。

The Limit Comparison Test says to make a ratio of the terms of two series and compute the limit. Unlike the Comparison Test, there is no need to compare the terms of both series. This test is most useful for series with rational expressions.
::限制比较测试表示要对两个序列的条款进行比对, 并计算限制。与比较测试不同, 没有必要对两个序列的条款进行比较。此测试对于使用理性表达方式的序列最为有用。

Let's apply the limit comparison test and determine if $\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{4} + 6 k^{3} - 1}{7 k^{5} + k^{2}} \end{aligned}$ converges or diverges.
::让我们应用限制比较测试, 确定“ k” = 1k4+6k3- 17k5+k2 是否趋同或不同。

Just as with rational functions, when $\begin{aligned} k \to \infty \end{aligned}$ the series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{4} + 6 k^{3} - 1}{7 k^{5} + k^{2}} \end{aligned}$ behaves like the series with only the highest powers of $\begin{aligned} k \end{aligned}$ in the numerator and denominator:
::和理性函数一样, 当 k 序列 =1\\\\ k4+6k3- 17k5+k2 表现得像序列时, k 在分子和分母中只有 k 的最大功率 :

\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{4}}{7 k^{5}} = \sum_{k = 1}^{\infty} \frac{1}{7 k} = \frac{1}{7} \sum_{k = 1}^{\infty} \frac{1}{k} . \end{aligned}

::1475117111111

We will use the series $\begin{aligned} \frac{1}{7} \sum_{k = 1}^{\infty} \frac{1}{k} \end{aligned}$ to apply the Limit Comparison Test.
::我们将使用17Qk=1Q1k序列来应用限制比较测试。

First, find the limit of the ratio of the terms of the two series:
::首先,找出两个系列条款比率的限度:

\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} & = lim_{k \to \infty} \frac{\frac{k^{4} + 6 k^{3} - 1}{7 k^{5} + k^{2}}}{\frac{1}{7 k}} \\ = lim_{k \to \infty} \frac{7 k^{4} + 42 k^{3} - 7}{7 k^{4} + k} \\ = 1 \end{aligned}

::立方公尺@ukvk=limkk4+6k3-17k5+k217k=limk7k4+42k3-77k4+k4=1

Since $\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} = 1 \end{aligned}$ , by the Limit Comparison Test, both $\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{4} + 6 k^{3} - 1}{7 k^{5} + k^{2}} \end{aligned}$ and $\begin{aligned} \frac{1}{7} \sum_{k = 1}^{\infty} \frac{1}{k} \end{aligned}$ either both converge or diverge.
::自limkukvk=1以来,根据限制比较测试, k=1k4+6k3-17k5+k2和17k=11k 两者均趋同或相左。

But, $\begin{aligned} \frac{1}{7} \sum_{k = 1}^{\infty} \frac{1}{k} \end{aligned}$ is a harmonic series, which is a series that diverges.
::但是,17k=11k 是一个调和序列, 它是一个相左的序列。

Therefore, $\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{4} + 6 k^{3} - 1}{7 k^{5} + k^{2}} \end{aligned}$ diverges.
::因此,k=1k4+6k3-17k5+k2的差异。

A simpler form of the Limit Comparison test, called the Simplified Limit Comparison Test , is as follows:
::称为简化的限制性比较试验的较简单形式如下:

Suppose $\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ are series with $\begin{aligned} u_{k} > 0 \end{aligned}$ and $\begin{aligned} v_{k} > 0 \end{aligned}$ for all $\begin{aligned} k \end{aligned}$ , and suppose
::假设“k”=1uk和“k”=1vk 与 uk>0 和 vk>0 的所有 k 序列,并假设

\begin{aligned} lim_{k \to \infty} \frac{u_{k}}{v_{k}} = L > 0, \end{aligned}

::立方公尺=L>0,

then either:
::然后要么:

$\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ both converge, or
::k=1 uk 和 k=1 vk 两者相趋近, 或

$\begin{aligned} \sum_{k = 1}^{\infty} u_{k} \end{aligned}$ and $\begin{aligned} \sum_{k = 1}^{\infty} v_{k} \end{aligned}$ both diverge.
::k=1uk 和 k=1vk 两者都不同。

Let's apply the Simplified Limit Comparison Test and determine if $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k} + 5} \end{aligned}$ converges or diverges.
::让我们应用简化限制比较测试, 确定“ k=1\\\\\\28k+5” 是否趋同或不同。

The series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k} + 5} \end{aligned}$ is a series without negative terms. We can apply the Simplified Limit Comparison Test by comparing the series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k} + 5} \end{aligned}$ with the series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k}} \end{aligned}$ which is a convergent geometric series.
::=k=128k+5序列是一个没有负条件的序列。我们可以将 &k=128k+5序列与Qk=128k+5序列(是一个可聚合的几何序列)进行比较,从而应用简化限制比较测试。

Then $\begin{aligned} lim_{k \to \infty} \frac{\frac{2}{8^{k} + 5}}{\frac{2}{8^{k}}} = lim_{k \to \infty} \frac{8^{k}}{8^{k} + 5} = 1 > 0 \end{aligned}$ .
::然后翻转28k+528k=limká8k8k+5=1>0。

Thus, since $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k}} \end{aligned}$ converges, then $\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{8^{k} + 5} \end{aligned}$ also converges.
::因此,既然=128k会合,那么k=128k+5也会汇合。

Examples
::实例

Example 1
::例1

Earlier, you were asked about the following two concepts:
::早些时候,有人问你们以下两个概念:

If a positive term series $\begin{aligned} \sum a_{n} \end{aligned}$ is always greater, term by term, than another positive term series $\begin{aligned} \sum b_{n} \end{aligned}$ that diverges, than the positive term series $\begin{aligned} \sum a_{n} \end{aligned}$ must also diverge;
::如果一个肯定术语系列在术语上总是大于另一个有差异的肯定术语系列,那么,肯定术语系列在术语上总是大于另一个有差异的肯定术语系列,那么肯定术语系列也必然存在差异;

If a positive term series $\begin{aligned} \sum a_{n} \end{aligned}$ is always smaller, term by term, than another infinite series $\begin{aligned} \sum b_{n} \end{aligned}$ that converges, than the positive term series $\begin{aligned} \sum a_{n} \end{aligned}$ must also converge.
::如果一个正值序列肯定值序列总是小, 逐个, 相对于另一个无穷无穷的序列 bn 相交合, 则肯定值序列肯定值序列也必须趋同。

If $\begin{aligned} △_{n} > 0 \end{aligned}$ represents how much greater or smaller a term $\begin{aligned} a_{n} \end{aligned}$ of a series is compared to the term $\begin{aligned} b_{n} \end{aligned}$ of a series of known convergence or divergence, can you formulate expressions that show (1) and (2)?
::如果“n>0”表示一个系列术语与一系列已知趋同或分歧的 bn 术语的大小,您能否写出显示(1)和(2)的表达式?

If $\begin{aligned} a_{n} - △_{n} = b_{n} \end{aligned}$ , then $\begin{aligned} \sum_{n = 1}^{\infty} (a_{n} - △_{n}) = \sum_{n = 1}^{\infty} b_{n} \Rightarrow \sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} b_{n} + \sum_{n = 1}^{\infty} △_{n} \end{aligned}$ . Since $\begin{aligned} \sum b_{n} \end{aligned}$ diverges, adding to its sum does not change divergence, and shows that $\begin{aligned} \sum a_{n} \end{aligned}$ must also be divergent.
::如果 ann=bn, 那么n=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

If $\begin{aligned} a_{n} + △_{n} = b_{n} \end{aligned}$ , then $\begin{aligned} \sum_{n = 1}^{\infty} (a_{n} + △_{n}) = \sum_{n = 1}^{\infty} b_{n} \end{aligned}$ . Since $\begin{aligned} \sum b_{n} \end{aligned}$ converges, $\begin{aligned} \sum_{n = 1}^{\infty} (a_{n} + △_{n}) = \sum_{n = 1}^{\infty} a_{n} + \sum_{n = 1}^{\infty} △_{n} \end{aligned}$ converges, which means both $\begin{aligned} \sum_{n = 1}^{\infty} a_{n} \end{aligned}$ and $\begin{aligned} \sum_{n = 1}^{\infty} △_{n} \end{aligned}$ must converge (the sum of a convergent and divergent series is divergent).
::如果ann=bn, 那么n=1( ann) n=1bn。自bn 集合,\n=1( ann) n=1n=1n=1n 集合, 这意味着n=1an 和n=1n 必须集合( 集合和不同序列的总和是不同的 ) 。

Example 2
::例2

Determine whether $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k} - 5} \end{aligned}$ converges or diverges.
::确定“ k” = 1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Notice that the series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k} - 5} \end{aligned}$ has some negative terms and can be written as
::请注意,Qk=111k4-5系列有某些负值,可以写成:

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k} - 5} \end{aligned}$
::k=11k4-5

The series $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k} - 5} \end{aligned}$ is similar to $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k}} = \sum_{k = 1}^{\infty} \frac{1}{k^{\frac{1}{4}}} \end{aligned}$ , which is a $\begin{aligned} p \end{aligned}$ -series, that diverges because $\begin{aligned} p < 1 \end{aligned}$ .
::“k”=1111k4-5序列与“k”=111k4}k=1111k14序列相似,该序列是一个p系列,因p<1而不同。

Using the Comparison Test, $\begin{aligned} \frac{1}{\sqrt[4]{k} - 5} \geq \frac{1}{\sqrt[4]{k}} \end{aligned}$ for all $\begin{aligned} k \end{aligned}$ .
::使用比较测试, 1k4 - 5Q1k4 适用于所有 k。

By the Comparison Test, $\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{\sqrt[4]{k} - 5} \end{aligned}$ also diverges.
::根据比较测试,k=11k4-5也存在差异。

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For #1-8, determine if the series converges or diverges using the Comparison Test.
::对于 # 1-8, 使用比较测试来确定序列是否趋同或不同。

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{n^{3} + n + 3} \end{aligned}$
::=1=11n3+n+3

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{2^{n} + 3} \end{aligned}$
::=1=1=12n+3

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{3^{n}}{5^{n} + 6} \end{aligned}$
::=1 =3n5n+6

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{2 n^{2} + n + 5} \end{aligned}$
::=1=1=12n2+n+5

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{(\sin n)^{2}}{n (n + 5)} \end{aligned}$
::Nn=1(sinn)2n(n+5)

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{1}{(4 k + 1)^{\frac{1}{2}}} \end{aligned}$
::*k=11( 4k+1)12

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{\arctan n}{2 n^{3}} \end{aligned}$
::=1 arctann2n3 =1 arctan_n2n3 =1 arctan_n2n3 =1 =1 arctan_n2n3

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{\sqrt{n + 4}}{n \sqrt{n + 2}} \end{aligned}$
::#n=1n+4nn+2

For #9-14, determine if the series converges or diverges using the Limit Comparison Test.
::对于# 9-14, 使用限制比较测试来确定序列是否趋同或不同。

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{2}{5 k^{5} - 4} \end{aligned}$
::=1=1 =25k5 -4

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{5}{(k + 1) (k + 3)} \end{aligned}$
::*k=15(k+1)(k+3)

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{k^{3} + 4 k^{2} + 1}{3 k^{6} + 2 k^{4}} \end{aligned}$
::k=1k3+4k2+13k6+2k4

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{4 k^{2} + 3 k + 9}{7 k^{3} + 11} \end{aligned}$
::k=14k2+3k+97k3+11

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{2 n^{3} + 4} \end{aligned}$
::Nn=1n2n3+4 =1n2n3+4

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{4^{n} + 5}{7^{n} + 13} \end{aligned}$
::=14n+57n+13

General Harmonic series $\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{a n + b} \end{aligned}$ with $\begin{aligned} a > 0 \end{aligned}$ .
::普通调力序列=11an+b, a>0。

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Section outline

General

阳性术语序列:比较测试

Comparison Tests ::比较测试

The (Direct) Comparison Test :直接)比较测试

The Limit Comparison Test ::限制性比较测试

Examples ::实例

Example 1 ::例1

Example 2 ::例2

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