9.10 肯定和负数系列:绝对和条件趋同

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  阳值和负值系列:绝对和条件趋同

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  In the previous concept, the was introduced as a type of infinite series with positive and negative terms, whose convergence could be determined by using the . However, not every infinite series with positive and negative terms is an alternating series, and for these the Alternating Series test is not applicable. For example, consider the series $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{\sin (nx)}{n}\end{array}$ . Is this an alternating series? Under what circumstances might it be a non-alternating series with positive and negative terms? Does it converge ?   
  ::在前一个概念中,引入了无穷无尽的系列,有正反两个词,两者的趋同可以通过使用来决定。然而,不是每个有正反两个词的无限系列都是交替的系列,对于这些系列来说,互换序列试验不适用。例如,考虑An=1sin(nx)n系列。这是一个交替系列吗?在什么情况下,它可能是带有正反两个词的非互换系列?它是否趋同?

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  Absolute and Conditional Convergence
  ::绝对和条件趋同

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  In the work with infinite series, the question has been whether the series converges or diverges . In this concept we present the fact that there are two types of convergence: absolute convergence and conditional convergence.
  ::在以无限系列开展工作的过程中,问题在于该系列是趋同还是不同,在这个概念中,我们提出有两种趋同:绝对趋同和有条件趋同。

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  Let $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k=u_1+u_2+u_3\dots +u_k+\dots \end{array}$  be an infinite series.
  ::将 & k=1 @ uk=u1+u2+u3...+uk+... 变成一个无限的序列 。

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  • The series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$  is absolutely convergent if the series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|\end{array}$  converges.
    ::“k”=1uk序列是绝对趋同的,如果“k”=1uk”序列是趋同的。

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  $$\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|=|u_1|+|u_2|+|u_3|+\dots +|u_k|+\dots \end{array}$$

  
  ::...

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  • The series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$  is conditionally convergent (or convergent) if it does converge but is not absolutely convergent.
    ::k=1uk 序列是有条件的集合( 或集合) , 如果它确实会趋同, 但不是绝对趋同 。

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  The infinite series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|\end{array}$ is a positive term series made by taking the absolute values of the terms of $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|\end{array}$ . This means that all the tools used to determine convergence for positive term series are applicable. What makes this important is that the convergence of the infinite sum of absolute values can tell us something about the convergence of the series.
  ::无限序列 Qk=1uk 是一个积极的术语序列, 以 Qk=1uk 术语的绝对值来计算。 这意味着用于确定正术语序列趋同的所有工具都适用。 使得这一点变得重要的是, 绝对值的无限和绝对值的趋同可以告诉我们关于此序列趋同的某种东西 。

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  Relationship between Absolute and Conditional Convergence
  ::绝对和条件趋同之间的关系

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  The relationship between absolute and conditional convergence is as follows:
  ::绝对趋同和有条件趋同之间的关系如下:

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  Let $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k=u_1+u_2+u_3\dots +u_k+\dots \end{array}$  be an infinite series.
  ::将 & k=1 @ uk=u1+u2+u3...+uk+... 变成一个无限的序列 。

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  1. If  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|\end{array}$ is absolutely convergent, then $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$  converges;
     ::如果“k”=1uk绝对趋同,那么“k”=1uk集合;
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  2. If  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|u_k|\end{array}$ is not absolutely convergent, then we cannot conclude anything about the convergence of $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$ .
     ::如果“k”=1uk不是绝对趋同,那么我们无法就“k”=1uk的趋同得出任何结论。

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  The above theorem tells us that if you can show absolute convergence, then the series converges. However, if the series of absolute value diverges, we cannot conclude anything about the series.
  ::上面的理论告诉我们,如果你能表现出绝对的趋同,那么系列就会趋同。然而,如果绝对值系列出现差异,我们就无法得出关于系列的任何结论。

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  Take series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{3k^4-k}\end{array}$ .  Let's  determine if it converges absolutely.
  ::使用 & k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ k+13k4\\\\\\\\\\\\\\\\\\\\\\\\\\\ k\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ &\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  This is an alternating series.
  ::这是一个交替系列。

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  We find the series of absolute values: $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|\frac{(-1{)}^{k+1}}{3k^4-k}|=\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k^4-k}\end{array}$ .
  ::我们发现绝对值的序列 :\ k=1 \ (- 1) k+13k4 - kQQ=1\\ 13k4- k) 。

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  Does the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k^4-k}\end{array}$ converge?
  ::序列 *k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Consider that $\begin{array}{r}0<\frac{1}{3k^4-k}<\frac{1}{3k^3}\end{array}$  for all $\begin{array}{r}k>1\end{array}$ . Because  $\begin{array}{r}\underset{k=2}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k^3}\end{array}$ is a $\begin{array}{r}p\end{array}$ -series with $\begin{array}{r}p=3\end{array}$ , it converges. Therefore the series  $\begin{array}{r}\underset{k=2}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k^4-k}\end{array}$ converges by the Comparison Test ; which also means that $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k^4-k}\end{array}$  converges.
  ::k>1. 由于 kk=2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  The series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{3k^4-k}\end{array}$ , therefore, converges absolutely and hence converges.
  ::因此, & k=1( - 1) k+13k4-k 序列会绝对趋同, 从而会趋同 。

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  Now, let's d etermine if the series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{2k+1}\end{array}$  converges absolutely.
  ::现在,让我们来决定该序列 'k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  This is an alternating series.
  ::这是一个交替系列。

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  First test for absolute convergence.
  ::对绝对趋同的第一次测试。

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  We find the series of absolute values: $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|\frac{(-1{)}^{k+1}}{2k+1}|=\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2k+1}\end{array}$ .
  ::我们找到绝对值序列 :\\ k=1\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\K12k+1\\\\\\\\\\\\\\\\\\\12K+1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Does the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2k+1}\end{array}$ converge?
  ::序列 *k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Consider that $\begin{array}{r}0<\frac{1}{3k}\le \frac{1}{2k+1}\end{array}$  for all $\begin{array}{r}k\end{array}$ . Because  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{3k}\end{array}$ is a harmonic series, it diverges. Therefore the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2k+1}\end{array}$ diverges by the Comparison Test.
  ::考虑所有 k 0< 13k12k+1。 因为 kk=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\> k+1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  The series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{2k+1}\end{array}$ , therefore, is not absolutely convergent.
  ::& k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1K+12k+1,所以,不是绝对的趋同。

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  Next test for conditional convergence (or simply convergence) by using the Alternating Series Test.
  ::利用交替系列试验进行有条件趋同(或简单趋同)的下一个试验。

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  First we check that the terms of the series are non-increasing: $\begin{array}{r}\frac{u_{k+1}}{u_k}\le 1\end{array}$ .
  ::首先,我们检查系列条款是否没有增加: uk+1uk1。

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  $$\begin{array}{r}\frac{u_{k+1}}{u_k}=\frac{\frac{1}{2(k+1)+1}}{\frac{1}{2k+1}}=\frac{2k+1}{2k+3}<1\end{array}$$

  
  ::uk+1uk=12(k+1)+112k+1=2k+12k+3 <1

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  Therefore, $\begin{array}{r}{u}_k>u_{k+1}\end{array}$ , and the first condition is satisfied.
  ::因此,uk>uk+1,满足第一个条件。

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  Next, we check that $\begin{array}{r}\underset{k\to +\mathrm{\infty }}{lim}{u}_k=0\end{array}$ .
  ::下一位,我们检查一下

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  $$\begin{array}{r}\underset{k\to +\mathrm{\infty }}{lim}{u}_k=\underset{k\to +\mathrm{\infty }}{lim}\frac{1}{2k+1}=0\end{array}$$

  
  ::立方公尺=立方公尺12k+1=0

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  By the Alternating Series Test, the series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{2k+1}\end{array}$  converges; it is conditionally convergent.
  ::在交替序列测试中,序列 'k=1\\\\\\\\\\\\1\K+12k+1"会合;是有条件的集合。

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  Rearrangement
  ::重新安排

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  Making a rearrangement of terms of a series means writing all of the terms of a series in a different order. The following theorem explains how rearrangement affects convergence.
  ::一系列条款的重新安排意味着将一系列条款的所有条款按不同的顺序写成。以下的理论解释重新安排如何影响趋同。

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  Rearrangement of Series Terms
  ::重新安排系列术语

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  1. Let  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$ is an absolutely convergent series. Then any rearrangement of terms in that series results in a new series that is also absolutely convergent to the same limit.
     ::Let 'k=1\uk 是一个绝对趋同的序列。 那么该序列中任何重新排列的术语都会产生一个新的序列, 并且绝对会达到相同的限制 。
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  2. Let  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{u}_k\end{array}$ be a conditionally convergent series. Then, for any real number $\begin{array}{r}c\end{array}$  there is a rearrangement of the series such that the new resulting series will converge to $\begin{array}{r}c\end{array}$ .
     ::将 & k=1 uk 变成一个有条件的集合序列。 然后, 对于任何真正数字 c , 都会对序列进行重新排列, 这样新的结果序列将合并到 c 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to determine if the  series $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{\sin (nx)}{n}\end{array}$  is an alternating series and whether or not it converges.
  ::早些时候,有人要求你确定序列“n”=1sin(nx)n是否是一个交替序列,以及该序列是否汇合。

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  It is not an alternating series. The nature of this series depends on the value of $\begin{array}{r}x\end{array}$ . If $\begin{array}{r}x=\frac{\pi }{2}\end{array}$ , the series is
  ::它不是一个交替序列序列。 此序列的性质取决于 x 的值。 如果 x\\\\ 2, 该序列是

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  $\begin{array}{r}1,0,-\frac{1}{3},0,\frac{1}{5},0,-\frac{1}{7},0,\dots \end{array}$  with an infinite number of positive and negative terms. This particular series could be written as a new alternating series  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}{\left(-\frac{1}{2n-1}\right)}^{n+1}\end{array}$ if the terms with values of 0 are deleted. Other values of $\begin{array}{r}x\end{array}$ , e.g., $\begin{array}{r}\frac{\pi }{4}\end{array}$ , would result in different positive and negative terms. The series $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{\sin (nx)}{n}\end{array}$  is convergent for every value of $\begin{array}{r}x\end{array}$ .
  ::1 0,-13,0,0,15,0,-17,0....,有无限数量的正值和负值。如果删除值为0的值值,此特定序列可以作为新的交替序列 'n=1(- 12n- 1)n+1 写入。其他x的值,例如 4,将产生不同的正值和负值。%n=1sin(nx)n 序列对于 x 的每一种值都是趋同的。

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  Example 2
  ::例2

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  Determine if the series $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{-3n}{n+1}\right)}^{4n}\end{array}$  converges absolutely.
  ::确定%k=1(- 3nn+1) 4n 序列是否绝对一致 。

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  We find the series of absolute values: $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}|{\left(\frac{-3n}{n+1}\right)}^{4n}|=\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{-3n}{n+1}\right)}^{4n}=\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{3n}{n+1}\right)}^{4n}\end{array}$ .
  ::我们发现绝对值的序列 :\k=1(- 3nn+1)4nk=1(- 3nn+1)4n}(- 3nn+1)4n}(- 3nn+1)4n=1(- 3nn+1)4n。

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  Notice that the use of absolute value for the series terms makes no difference on the series because $\begin{array}{r}(-1{)}^{4n}=1\end{array}$ .
  ::请注意,使用序列术语的绝对值对序列没有区别,因为(-1)4n=1。

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  Does the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{3n}{n+1}\right)}^{4n}\end{array}$ converge?
  ::序列 'k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\N4n集中?

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  Since the series involves a power of $\begin{array}{r}n\end{array}$ , let’s try using the .
  ::因为这系列涉及n的力量,

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  $$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{a_n}& =\underset{n\to \mathrm{\infty }}{lim}{\left({\left(\frac{3n}{n+1}\right)}^{4n}\right)}^{\frac{1}{n}}\\ & =\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3n}{n+1}\right)}^4\\ & =\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3}{1+\frac{1}{n}}\right)}^4\\ & =3^4>1\end{array}$$

  
  ::limnann=limn (( 3nn+114n) 1n=limn (3nn+1) 4=limn( 31+1n) 4=34>1

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  The fact that $\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{a_n}>1\end{array}$  means that the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{3n}{n+1}\right)}^{4n}\end{array}$ diverges.
  ::Limnann>1 表示序列Qk=1( 3nn+1) 4 存在差异。

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  Example 3
  ::例3

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  Determine if $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{k}\end{array}$  converges absolutely, converges conditionally, or diverges.
  ::确定 & k=1\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1K+1k\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  To test for absolute convergence, look at the series of absolute values: $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{k}\end{array}$ . This is the harmonic series, which diverges. So, the series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{k}\end{array}$ does not converge absolutely.
  ::要测试绝对趋同性, 请查看绝对值序列 :\\ k= 1\\\\\ k。 这是调和性序列, 不同 。 因此, \k= 1\\\\\\\\\\\\\\\\\\\\\ k+1k 序列并非绝对一致 。

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  The next step is to check the convergence of $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{k}\end{array}$ . This will tell us if the series converges conditionally.
  ::下一步是检查 & k=1\\\\\\\\\\\\\\\1\k+1k\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Applying the Alternating Series Test:
  ::应用交替序列测试 :

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  The $\begin{array}{r}\frac{1}{1}>\frac{1}{2}>\frac{1}{3}>\dots \end{array}$  is non increasing and $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim}\frac{1}{k}=0\end{array}$ .
  ::11>12>13>... 不增加, 和limk1k=0。

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  The series  $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{k}\end{array}$ converges.
  ::& k=1\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ k+1k\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Hence, the series converges conditionally, but not absolutely.
  ::因此,该系列是有条件的,但不是绝对的。

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  Review
  ::回顾

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  Determine if each series converges absolutely, converges conditionally, or diverges.
  ::确定每个序列是否绝对趋同、有条件趋同或有分歧。

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  1. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{n+1}}{n^{\frac{1}{4}}}\end{array}$
     ::n=1(- 1)n+1n14
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  2. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^n}{n^4}\end{array}$
     ::n=1(- 1)nn4
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  3. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^{k+1}\frac{3k}{2^k}\end{array}$
     ::*k=1(- 1) k+13k2k
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  4. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}k}{2k^2+2}\end{array}$
     ::*k=1(- 1) k+1k2k2+2
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  5. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-4{)}^{k+1}}{7k^2}\end{array}$
     ::k=1(- 4) k+17k2
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  6. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^{k+1}}{k^{\frac{7}{2}}}\end{array}$
     ::*k=1(- 1) k+1k72
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  7. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^{k+1}\frac{k}{3k^2+k}\end{array}$
     ::*k=1(- 1,k+1k3k2+k)
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  8. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^n{e}^{\frac{1}{n}}}{n^3}\end{array}$
     ::n=1(- 1) nne1nn3
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  9. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^{k+1}\frac{1}{n\sqrt{n}}\end{array}$
     ::*k=1(- 1) k+11nn
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  10. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^{n+1}\frac{n}{e^n}\end{array}$
      ::n=1(- 1)n+1nen
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  11. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-2{)}^n}{n^2}\end{array}$
      ::=1 (-2)nn2
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  12. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^n\frac{n}{(n+1)\sqrt{n}}\end{array}$
      ::n=1(- 1)nn( n+1)n
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  13. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{(-1{)}^n}{n\sqrt{n^2+2}}\end{array}$
      ::n=1(- 1)nn2+2
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  14. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^n\frac{2n^2}{n^3+1}\end{array}$
      ::n=1(- 1)n2n2n2n3+1
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  15. $\begin{array}{r}\underset{k=1}{\overset{\mathrm{\infty }}{\sum }}(-1{)}^k\frac{\ln k}{\sqrt{k}}\end{array}$
      ::k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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