9.12 趋同试验摘要

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  趋同试验摘要

  As a preview of the later concepts, suppose you are given the infinite series $\begin{align*}\sum\limits_{n=1}^\infty 1.5 x^{n-1}\end{align*}$ , where  $\begin{align*}x\end{align*}$ is a real variable, and told that it can be a positive term series, or an , and that it can converge absolutely or diverge . In addition, you are told that this infinite series can be the representation of a function $\begin{align*}f(x)\end{align*}$ . Using the concepts you have learned about infinite series, can you determine whether these statements are true or false? 

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  Summary of Convergence Tests
  ::趋同试验摘要

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  We have seen various tests for determining the convergence of an infinite series. It is sometimes difficult to choose the best convergence test for a particular series. Not all tests work on any given series. Multiple tests may work on a given series, but even if a test works on a particular series, that test may still involve a lot of work in reaching a convergence conclusion. With experience, however, we can learn to apply the right test to a given series. At a minimum, we can learn which tests are easiest to apply, so that we can start with those easier tests if we have no other idea how proceed. A key factor is to be able to recognize as many forms as possible.
  ::我们看到了用来确定无限序列趋同的各种测试。 有时很难为特定序列选择最佳趋同测试。 不是所有测试都针对任何特定序列。 多个测试可能针对一个特定序列起作用, 但即使对一个特定序列进行测试, 该测试仍可能涉及许多工作来得出趋同结论。 但是,根据经验, 我们可以学会对一个特定序列应用正确的测试。 我们至少可以知道哪些测试最容易应用, 以便我们能够从这些比较简单的测试开始, 如果我们没有其它的操作方法。 一个关键因素是能够识别尽可能多的形式。

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  In any case, when you are testing a series for convergence or divergence, it's helpful to have a plan of attack. Ask the following questions and then select appropriate tests:
  ::无论如何,当您在测试一系列的趋同或分歧时, 制定攻击计划是有帮助的。 问下列问题然后选择适当的试验:

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  1. Is the series a recognizable series type with known convergence characteristics? If so look at the available test options. If not, then ask the next question.
     ::序列是否为具有已知趋同特性的可识别序列类型?如果是的话,请查看现有的测试选项。如果没有,请问下一个问题。
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  2. Is the series a positive term series? If so look at the available test options. If not, then ask the next question.
     ::序列是否为正术语序列? 如果是的话, 请查看可用的测试选项。 如果没有, 请问下一个问题 。
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  3. Is the series a positive and negative term series? If so look at the available test options.
     ::系列是正值和负值系列吗?如果是的话,请看看现有的测试选项。

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  The following is a framework for making test choices. It provides a summary of the various tests and when they might be useful.
  ::以下是作出试验选择的框架,概述了各种试验及其可能有用的时间。

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  I. The series is recognizable with known convergence characteristics.
  ::一. 该系列具有已知的趋同特点。

  Recognizable Infinite Series

  播放段落 Type of Series ::系列类型	播放段落 Convergence/Divergence Test ::趋同/活力试验
  播放段落 A recognizable series type : ::可识别序列类型: 播放段落 Geometric : $\begin{align*}\sum\limits_{n=1}^\infty ar^{n-1}\end{align*}$ ::几何: @ n=1arn- 1	播放段落 Geometric series test:  $\begin{align*}|r|<1\Rightarrow\end{align*}$ convergence ::几何序列试验: 播放段落 $\begin{align*}|r|\ge 1\Rightarrow\end{align*}$ divergence ::差异
  播放段落 Harmonic : $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{n}\end{align*}$ ::调音: @n=11n	播放段落 Diverge ::下游
  播放段落 $\begin{align*}p\end{align*}$ -series : $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{n^{p}}\end{align*}$ ::p- 系列: @n=11np	播放段落 $\begin{align*}p\end{align*}$ -series Test:  $\begin{align*}p>1\Rightarrow\end{align*}$ convergence ::p系列试验:p>1_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 播放段落 $\begin{align*}0 < p \le 1\Rightarrow\end{align*}$ divergence ::0<p1\\\\\\\ 差异
  播放段落 Telescoping : $\begin{align*}\sum\limits_{k=1}^\infty(a_k - a_{k+1})\end{align*}$ ::遥测范围: k=1( ak- ak+1)	播放段落 $\begin{align*}\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(a_1 - a_n)=\text{finite}\Rightarrow\end{align*}$ convergence :a1 - an) = finite = 趋同 播放段落 $\begin{align*}\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(a_1 - a_n)=\pm \infty\end{align*}$  or DNE $\begin{align*}\Rightarrow\end{align*}$ divergence ::=====================================================================================================================================================================================================================

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  Let's use the information above to determine whether the each of the following series is convergent, or divergent.
  ::让我们利用上述信息来决定下一系列中的每一系列是趋同的还是不同的。

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  • 1. $\begin{align*}\sum\limits_{k=1}^\infty \frac{4^{k+2}}{9^{k-1}}\end{align*}$
    ::1. *k=14k+29k-1
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  • 2. $\begin{align*}\sum\limits_{k=1}^\infty\frac{5}{(2k)^{\frac{2}{3}}}\end{align*}$
    ::2. *k=15(2k)23

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  $\begin{align*}\sum\limits_{k=1}^\infty\frac{4^{k+2}}{9^{k-1}}=\sum\limits_{k=1}^\infty 4^{3}\left(\frac{4}{9}\right)^{k-1}\end{align*}$ is a geometric series with $\begin{align*}a=4^{3}\end{align*}$  and $\begin{align*}r=\frac{4}{9}\end{align*}$ . Because $\begin{align*}|r|=\frac{4}{9}<1\end{align*}$ , the series is convergent.
  ::k=14k+29k- 1k=143(49k- 1)是一个几何序列, a=43 和 r=49。 因为 @r=49 < 1, 该序列是集合的 。

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  $\begin{align*}\sum\limits_{k=1}^\infty\frac{5}{(2k)^{\frac{2}{3}}}=\sum\limits_{k=1}^\infty\frac{5}{2^{\frac{2}{3}}}\frac{1}{k^{\frac{2}{3}}}=\frac{5}{2^{\frac{2}{3}}}\sum\limits_{k=1}^\infty\frac{1}{k^{\frac{2}{3}}}\end{align*}$ . This is a $\begin{align*}p\end{align*}$ -series with $\begin{align*}p=\frac{2}{3}\end{align*}$ . Because $\begin{align*}p=\frac{2}{3}<1\end{align*}$ , the series diverges .
  ::@k=15( 2k)23\k=152231K23=5223\k=11k23。 这是一个 p=23的p系列。 因为p=23<1, 序列会有所不同 。

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  II. The series is a positive term series.
  ::二. 该系列是一个积极的术语系列。

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  A Positive Term Series : $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  with $\begin{align*}u_k\ge 0\end{align*}$
  ::正数序列数: @k=1uk 和 uk_0

  播放段落 Test ::测试测试测试	播放段落 Comment ::注释注释注释注释
  播放段落 1. $\begin{align*}n\end{align*}$ th term divergence Test ::1. n n 术语差数测试 播放段落 $\begin{align*}\lim\limits_{k \to +\infty}u_k\ne 0\end{align*}$  or DNE  $\begin{align*}\Rightarrow\end{align*}$ divergence ::limkuk0 或 DNE 差异 播放段落 $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\Rightarrow\end{align*}$  no conclusion; try another test ::limkuk=0\\\\\\\无结论; 尝试另一个测试	$\begin{align*}f(k)\end{align*}$  is continuous, positive, decreasing and easily integrable.
  播放段落 2. ( $\begin{align*}u_k=f(k); f(k)\end{align*}$ is continuous, positive, decreasing and easily integrable) ::2. (uk=f(k);f(k) 是连续的、正的、递减的和容易辨认的) 播放段落 $\begin{align*}\int\limits_{1}^{\infty}f(x)dx=L, 0 < L < \infty\Rightarrow\end{align*}$  convergence ::=1f(x)dx=L,0<L 趋同 播放段落 $\begin{align*}\int\limits_{1}^{\infty}f(x)dx=\infty\Rightarrow\end{align*}$ divergence ::1f(x)dx	播放段落 $\begin{align*}f(k)\end{align*}$  is continuous, positive, decreasing and easily integrable. ::f(k) 是连续的、积极的、不断减少的和易于辨认的。
  播放段落 3. Comparison Test (with $\begin{align*}\sum\limits_{k=1}^\infty v_k, v_k\ge 0\end{align*}$ ) ::3. 比较测试(与k=1vk,vk0) 播放段落 $\begin{align*}u_k \le v_k\end{align*}$ and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$   converges $\begin{align*}\Rightarrow \sum\limits_{k=1}^\infty u_k\end{align*}$  convergence ::ukvk 和 k=1vk 趋同 *k=1uk 趋同 播放段落 $\begin{align*}u_k \ge v_k\end{align*}$ and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$  diverges, then $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  divergence ::ukvk 和 {k=1vk 差异,然后 @k=1uk 差异	$\begin{align*}f(k)\end{align*}$  is continuous, positive, decreasing and easily integrable.
  播放段落 4. Limit Comparison Test (with $\begin{align*}\sum\limits_{k=1}^\infty v_k, v_k\ge 0\end{align*}$ ) ::4. 限制比较测试(使用k=1vk,vk0) 播放段落 Compute $\begin{align*}\lim\limits_{k \to \infty}\frac{u_k}{v_k}=L\end{align*}$ ::计算 limkukvk=L 播放段落 $\begin{align*}0 < L < \infty\Rightarrow\end{align*}$ both series converge or diverge together. ::0<L两个系列相趋近或相距不一。 播放段落 $\begin{align*}L=0\end{align*}$ and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$  converges $\begin{align*}\Rightarrow\sum\limits_{k=1}^\infty u_k\end{align*}$  converges ::L=0 和 k=1vk 趋同 k=1uk 趋同 播放段落 $\begin{align*}L=+\infty\end{align*}$ and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$  diverges $\begin{align*}\Rightarrow\sum\limits_{k=1}^\infty u_k\end{align*}$  diverges ::LQQ 和 @k=1vk 差数 @k=1uk 差数	$\begin{align*}f(k)\end{align*}$  is continuous, positive, decreasing and easily integrable.
  播放段落 5. - compute $\begin{align*}\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_n}=L\end{align*}$ ::5. 计算limnan+1an=L 播放段落 $\begin{align*}L<1 \Rightarrow\end{align*}$  convergence ::L<1__+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++%++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 播放段落 $\begin{align*}L>1\Rightarrow\end{align*}$ divergence ::L>1差异 播放段落 $\begin{align*}L=1\Rightarrow\end{align*}$ not conclusive; try another test ::L=1\\\\\\ 尚未确定; 尝试另一个测试	Use Ratio Test when $\begin{align*}n\end{align*}$ th term involves factorials or $\begin{align*}n\end{align*}$ th powers. Avoid the Ratio Test for $\begin{align*}n\end{align*}$ th terms containing a rational expression involving only polynomials or polynomials under radicals, because the result will always be $\begin{align*}L=1\end{align*}$ .
  播放段落 6. - compute $\begin{align*}\lim\limits_{n \to \infty}\sqrt[n]{a_n}=L\end{align*}$ ::6. 计算limnnan=L 播放段落 $\begin{align*}L<1 \Rightarrow\end{align*}$  convergence ::L<1__+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++%++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 播放段落 $\begin{align*}L>1 \Rightarrow\end{align*}$ divergence ::L>1差异 播放段落 $\begin{align*}L=1\Rightarrow\end{align*}$ not conclusive; try another test ::L=1\\\\\\ 尚未确定; 尝试另一个测试	Use the Root Test if the $\begin{align*}n\end{align*}$ th term contains products or quotients of powers of $\begin{align*}n\end{align*}$ .

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  Let's use the information above to determine whether the each of the following series is convergent, or divergent.
  ::让我们利用上述信息来决定下一系列中的每一系列是趋同的还是不同的。

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  • $\begin{align*}\sum\limits_{n=1}^\infty\frac{n2^{n}}{n!}\end{align*}$
    ::1n2nn! 10n2n! 10nn! 10nn! 10nn! 10nn!
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  • $\begin{align*}\sum\limits_{n=2}^\infty\frac{1}{n(\ln n)}\end{align*}$
    ::*n=21n(lnn) *n=21n(lnn)

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  $\begin{align*}\sum\limits_{n=1}^\infty\frac{n2^{n}}{n!}=\sum\limits_{n=1}^\infty\frac{n2\cdot 2^{n-1}}{n!}=2\sum\limits_{n=1}^\infty\frac{2^{n-1}}{(n-1)!}\end{align*}$  is a positive term series. Since the series has a factorial, try the Ratio Test.
  ::n=1n2n2n_1n!=2n=12n_1(n_1)!这是一个正术语序列。由于该序列有一个因数,请尝试比率测试 。

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  !%7D%7D%7B%5Cfrac%7B2%5E%7Bn-1%7D%7D%7B(n-1)!%7D%7D%5C%5C%0A%26%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%5E%7Bn%7D%7D%7B2%5E%7Bn-1%7D%7D%5Cfrac%7B(n-1)!%7D%7B!%7D%5C%5C%0A%26%20%3D%20%5Clim%5Climits_%7Bn%20%5Cto%20%5Cinfty%7D2%20%5Cfrac%7B1%7D%7Bn%7D%5C%5C%0A%26%20%3D%200%3C1%0A">

  $$\begin{align*}\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_n} & = \lim\limits_{n \to \infty}\frac{\frac{2^{n}}{(n)!}}{\frac{2^{n-1}}{(n-1)!}}\\ & = \lim_{n \to \infty}\frac{2^{n}}{2^{n-1}}\frac{(n-1)!}{(n)!}\\ & = \lim\limits_{n \to \infty}2 \frac{1}{n}\\ & = 0<1 \end{align*}$$
  ::========================================================================================================================================================================================================================================================================================================================================================================================

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  The series is absolutely convergent .
  ::这系列绝对是趋同的。

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  The series  $\begin{align*}\sum\limits_{n=2}^\infty\frac{1}{n(\ln n)}\end{align*}$ is a positive term series.
  ::=n=21n(lnn)序列是一个正数术语序列 。

  播放段落

  The test for $\begin{align*}n\end{align*}$ th term divergence is inconclusive because $\begin{align*}\lim\limits_{n \to \infty}\frac{1}{n(\ln n)}=0\end{align*}$ .
  ::nth 条件差的测试没有结论, 因为 limn1n( lnn) =0 。

  播放段落

  The  $\begin{align*}n\end{align*}$ th term looks integrable, so try the Integral Test: let $\begin{align*}f(x)=\frac{1}{x(\ln x)},x\ge 2\end{align*}$ .
  ::nth 术语看起来是不可解码的, 所以尝试综合测试 : 让 f( x) = 1x( lnx) , x% 2 。

  播放段落

  $$\begin{align*}\int\limits_{2}^{\infty}f(x)dx & = \lim\limits_{p \to \infty}\int\limits_{2}^{p}\frac{1}{x(\ln x)}dx \qquad \ldots\text{Let} \ u=\ln x,du=\frac{dx}{x}\\ & = \lim\limits_{p \to \infty}\int\limits_{\ln 2}^{\ln p}\frac{1}{u}du\\ & = \lim\limits_{p \to \infty}[\ln(\ln p)-\ln(\ln 2)]\\ & = \infty\end{align*}$$
  ::2f( x) dx=limpp21x( lnx) dx... Let u=lnx, du=dxx=limplnp ln21udu=limp[ln( lnp)- ln( ln2)] {

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  The series $\begin{align*}\sum\limits_{n=2}^\infty\frac{1}{n(\ln n)}\end{align*}$  diverges by the Integral Test.
  ::综合测试的序列 = 21n( lnn) 差异 。

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  III. The series is a positive and negative term series
  ::三. 该系列是正和负术语系列

  A Positive & Negative Term Series: Alternating  & non-Alternating Series

  播放段落 Type of Series ::系列类型	播放段落 Test ::测试测试测试
  播放段落 Alternating or non-alternating series ::交替或非交替序列号	播放段落 Test for absolute convergence or divergence ::绝对趋同或分歧的试验 播放段落 Absolute Convergence Test : ::绝对趋同测试 : 播放段落 Form the new series $\begin{align*}\sum|u_k|\end{align*}$  or  $\begin{align*}\sum\left|(-1)^{k-1}u_k\right|\end{align*}$ and evaluate using the positive term series tests: ::以新序列 {uk} 或\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\-\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\用正术语序列测试来评估 使用正词序列测试 : 播放段落 Convergence $\begin{align*}\Rightarrow\end{align*}$  absolute convergence $\begin{align*}\Rightarrow\end{align*}$  original series converges. ::趋同 绝对趋同 原始序列趋同 播放段落 Divergence $\begin{align*}\Rightarrow\end{align*}$  absolute divergence $\begin{align*}\Rightarrow\end{align*}$  original series may converge (conditionally) or diverge. ::原系列可能(有条件)趋同或不同。 播放段落 Inconclusive  $\begin{align*}\Rightarrow\end{align*}$ try another test. ::没有结论的 尝试另一个测试。 播放段落 $\begin{align*}n\end{align*}$ -th term divergence Test ::n- 第 n 术语差数测试 播放段落 $\begin{align*}\lim\limits_{k \to +\infty}u_k\ne0\end{align*}$  or DNE  $\begin{align*}\Rightarrow\end{align*}$ absolute divergence  $\begin{align*}\Rightarrow\end{align*}$ original series diverges. ::原始序列有差异。 播放段落 Ratio Test - compute $\begin{align*}\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=L\end{align*}$ ::比率测试 - 计算limnan+1anL 播放段落 $\begin{align*}L<1 \Rightarrow\end{align*}$  convergence ::L<1__+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++%++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 播放段落 $\begin{align*}L>1\Rightarrow\end{align*}$ divergence ::L>1差异 播放段落 $\begin{align*}L=1\Rightarrow\end{align*}$ not conclusive; try another test ::L=1\\\\\\ 尚未确定; 尝试另一个测试
  播放段落 Alternating Series : $\begin{align*}\sum\limits_{k=1}^\infty(-1)^{k-1}u_k\end{align*}$ ::交换序列 : @k=1(- 1)- 1uk 播放段落 or $\begin{align*}\sum\limits_{k=1}^\infty(-1)^{k}u_k\end{align*}$ with $\begin{align*}u_k>0\end{align*}$  for all $\begin{align*}k\end{align*}$ ::或 kk=1 (- 1) 或 kk=1 (- 1) , 有 uk>0 的 kuk 和 uk>0 。	播放段落 for (conditional) convergence :有条件)趋同 播放段落 $\begin{align*}u_k\ge u_{k+1}\end{align*}$ (for all $\begin{align*}k\end{align*}$ ) and $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\Rightarrow\end{align*}$  convergence; ::ukuk+1 (所有 k) 和limkuk=0 趋同; 播放段落 Note : The failure to meet one or both of the conditions $\begin{align*}u_k\ge u_{k+1}\end{align*}$ or  $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\end{align*}$   does not mean that the series diverges. Another test must be tried, e.g., the $\begin{align*}n\end{align*}$ th term divergence test. ::注意: 无法满足 ukuk+1 或 limkuk=0 两种条件中的一种或两种条件,并不意味着序列有差异。必须尝试另一个测试,例如 nth 术语差异测试。

  播放段落

  Let's use the information above to determine whether the series $\begin{align*}\sum\frac{(-1)^{n-1}}{n^{q}}\end{align*}$ , where $\begin{align*}q\end{align*}$  is a real number, is absolutely convergent, conditionally convergent , or divergent.
  ::让我们使用上面的信息来确定序列 Q(-1)n-1nq,其中q是一个真实数字, 是绝对趋同的, 有条件的趋同的, 还是不同的。

  播放段落

  The series  $\begin{align*}\sum\frac{(-1)^{n-1}}{n^{q}}\end{align*}$ is an alternating series. Does it satisfy the necessary conditions for convergence?
  ::*(-1)n-1nq系列是一个交替序列。它是否满足了趋同的必要条件?

  播放段落

  The $\begin{align*}n\end{align*}$ th term  $\begin{align*}\frac{1}{n^{q}}\end{align*}$ is decreasing and $\begin{align*}\lim\limits_{n \to \infty}\frac{1}{n^{q}}=0\end{align*}$  only for $\begin{align*}q>0\end{align*}$ .
  ::Nth 术语 1nq 正在减少, 只有 q> 0 的 limn1nq=0 。

  播放段落

  The series converges for $\begin{align*}q>0\end{align*}$ . Does it converge conditionally or absolutely?
  ::Q>0 的序列会合并。 它会有条件还是绝对会合并 ?

  播放段落

  The series $\begin{align*}\sum\limits_{n=1}^\infty\left|\frac{(-1)^{n-1}}{n^q}\right|=\sum\limits_{n=1}^\infty\frac{1}{n^q}\end{align*}$  is a $\begin{align*}p\end{align*}$ -series which converges for $\begin{align*}q>1\end{align*}$ .
  ::==1(- 1)-1-nqn=11nq 序列是一个为 q>1 集合的p系列。

  播放段落

  Therefore, if $\begin{align*}q>1\end{align*}$  the series $\begin{align*}\sum\frac{(-1)^{n-1}}{n^q}\end{align*}$  converges absolutely.
  ::因此,如果 q>1 序列 (-1)n-1nq 绝对一致。

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  For $\begin{align*}0 < q \le 1\end{align*}$ , the series $\begin{align*}\sum\frac{(-1)^{n-1}}{n^q}\end{align*}$  converges conditionally.
  ::0<q1, 序列 (- 1)n- 1nq 有条件交汇 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked whether the following statements about the infinite series $\begin{align*}\sum\limits_{n=1}^\infty 1.5 x^{n-1}\end{align*}$ , where  $\begin{align*}x\end{align*}$ is a real variable, are true or false.
  ::早些时候,有人问您,以下关于无限序列的描述是否真实或虚假? @n=11.5xn-1,其中x是一个真正的变量。

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  1. the series can be a positive term series, or an alternating series
     ::或交替序列
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  2. the series can converge absolutely or diverge
     ::该序列可以绝对或相异地对齐
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  3. the series can be the representation of a function $\begin{align*}f(x)\end{align*}$ .
     ::序列可以是函数 f(x) 的表示。

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  Were you able to come up with these answers?
  ::你能想出这些答案吗?

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  1. True: the series can be a positive term series for $\begin{align*}x>0\end{align*}$ , or an alternating series for $\begin{align*}x<0\end{align*}$ .
     ::真理: 该序列可以是 x>0 的正值术语序列, 也可以是 x<0 的交替序列 。
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  2. True: the series is a geometric series $\begin{align*}\sum\limits_{n=1}^\infty ar^{n-1}\end{align*}$  with $\begin{align*}a=1.5\end{align*}$  and $\begin{align*}r=x\end{align*}$  that will converge absolutely when $\begin{align*}|x|<1\end{align*}$ , and diverge when $\begin{align*}|x|\ge1\end{align*}$ .
     ::校对:该序列是一个几何序列 *n=1arn-1, 带有 a=1.5 和 r=x, 当\\x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  3. True: For $\begin{align*}|x|<1\end{align*}$  where the series converges absolutely,  $\begin{align*}\sum\limits_{n=1}^\infty 1.5 x^{n-1}=\frac{1.5}{1-x}=f(x)\end{align*}$ .
     ::True: 对于序列绝对相近的 x1,\\ n= 11.5xn- 1=1.51-x=f(x) 。

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  The series $\begin{align*}\sum\limits_{n=1}^\infty 1.5 x^{n-1}\end{align*}$  is a very important type of series called a power series , and you have the tools to be able to see how useful it can be! 
  ::“n”=11.5xn-1系列是一个非常重要的系列类型,叫做电源序列,您有工具可以看它有多有用!

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  Example 2
  ::例2

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  Determine whether the series $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}\sqrt{n}}{n+1}\end{align*}$  is absolutely convergent, conditionally convergent, or divergent.
  ::确定序列 *n=1(-1)n- 1nn+1是绝对趋同、有条件趋同还是相左。

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  The series $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}\sqrt{n}}{n+1}\end{align*}$  is an alternating series. Does it satisfy the necessary conditions for convergence?
  ::=n=1(- 1)n- 1n+1系列是一个交替序列。它是否满足了趋同的必要条件?

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  The $\begin{align*}n\end{align*}$ th term  $\begin{align*}\frac{\sqrt{n}}{n+1}\end{align*}$ is decreasing (look at the derivative of $\begin{align*}\frac{\sqrt{x}}{x+1}\end{align*}$ ), and $\begin{align*}\lim\limits_{n \to \infty}\frac{\sqrt{n}}{n+1}=\lim\limits_{n \to \infty}\frac{\frac{1}{\left (2\sqrt{n} \right )}}{1}=0\end{align*}$ . The alternating series converges. Does it converge conditionally or absolutely?
  ::nth unterm = nn+1 正在递减( 查看 xx+1 的衍生物) , 和 limn}nn+1 =limn}1 (2n)1 =0。 交替序列会合并 。 它有条件还是绝对会合并 ?

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  To determine whether the series is absolutely convergent we look at the series $\begin{align*}\sum\limits_{n=1}^\infty\left|\frac{(-1)^{n-1}\sqrt{n}}{n+1}\right|=\sum\limits_{n=1}^\infty \frac{\sqrt{n}}{n+1}\end{align*}$ .
  ::为了确定该序列是否绝对趋同,我们查看该序列 *n=1(-1)n-1nn+1n=1nn+1)。

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  Notice that $\begin{align*}\frac{\sqrt{n}}{n+1}\ge \frac{1}{n+1}\end{align*}$ , so that $\begin{align*}\sum\limits_{n=1}^\infty\frac{\sqrt{n}}{n+1}\ge\sum\limits_{n=1}^\infty\frac{1}{n+1}\end{align*}$ .
  ::注意\nn+11n+1, 所以\n=1\n+1\n=11n+1。

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  By the Integral Test, $\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{n+1}\end{align*}$  diverges. This means that by the Comparison Test, the series $\begin{align*}\sum\limits_{n=1}^\infty\frac{\sqrt{n}}{n+1}\end{align*}$  also diverges.
  ::在综合测试中,=n=11n+1的差异。这意味着在比较测试中,=n=1n+1的序列也不同。

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  Therefore the series $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}\sqrt{n}}{n+1}\end{align*}$  is not absolutely convergent, but it is conditionally convergent.
  ::因此, =n=1(-1)n- 1nn+1的序列并非绝对趋同,而是有条件趋同。

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  Review
  ::回顾

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  For each of the series, determine if it is absolutely convergent, conditionally convergent, or divergent.
  ::对于每一系列,确定它是否绝对趋同、有条件趋同或不同。

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  1. $\begin{align*}\sum\limits_{n=3}^\infty\frac{5}{n-2}\end{align*}$  
     ::#n=35n-2# #n=35n-2#
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  2. $\begin{align*}\sum\limits_{n=2}^\infty\frac{\ln n}{n^2} \end{align*}$
     ::*n=2nnn2
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  3. $\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{2n^{2}+n+5}\end{align*}$
     ::*n=112n2+n+5
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  4. $\begin{align*}\sum\limits_{k=1}^\infty\frac{k^{3}+4k^{2}+1}{3k^{6}+2k^{4}}\end{align*}$
     ::k=1k3+4k2+13k6+2k4
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  5. $\begin{align*}\sum\limits_{n=1}^\infty\frac{n!}{n^n}\end{align*}$
     ::不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不!
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  6. $\begin{align*}\sum\limits_{n=1}^\infty n\left(\frac{2}{3}\right)^n\end{align*}$
     ::n=1n( 23n)
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  7. $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-2)^{n}n}{7^{n+1}}\end{align*}$
     ::n=1(-2)nn7n+1
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  8. $\begin{align*}\sum\limits_{n=1}^\infty(-1)^n\frac{3n^{4}+5}{4n^{9}-3n}\end{align*}$
     ::*n=1(- 1)n3n4+54n9-3n
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  9. $\begin{align*}\sum\limits_{n=2}^\infty\frac{(-1)^n}{\sqrt{n^{2}-1}}\end{align*}$
     ::*n=2(-1)n_n2-1
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  10. $\begin{align*}\sum\limits_{n=1}^\infty\frac{\cos n}{n^3}\end{align*}$
      ::N=1consnn3 = 1consnn3 = 1consnn3 = 1consnn3 = 1consnn3 = 1consnn3
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  11. $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^n}{\sqrt[3]{n^2}}\end{align*}$
      ::n=1(- 1) n3n2
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  12. $\begin{align*}\sum\limits_{n=1}^\infty\frac{\sin n}{n^2}\end{align*}$
      ::*n=1sinnn2
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  13. $\begin{align*}\sum\limits_{n=2}^\infty\frac{(-1)^n}{\ln n}\end{align*}$
      ::n=2 (- 1) nlnn
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  14. $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^{n}2^n}{2n}\end{align*}$
      ::*n=1(- 1)n2n2nn
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  15. $\begin{align*}\sum\limits_{n=1}^\infty\frac{(-1)^{n}n!}{3^n}\end{align*}$
      ::N=1 (-1) nn! 3n!

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  Review (Answers)
  ::回顾(答复)

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  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 。