9.9 肯定和否定术语:替代术语系列

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  阳性和负值术语:

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  Not all infinite series are positive term series. Some contain positive and negative terms. Consider the series $\begin{align*}\sum\limits_{n=1}^\infty 8 \left(-\frac{1}{2}\right)^{n-1}\end{align*}$ , whose terms alternate in sign. What type of series this is? How would you determine whether it converges or diverges ?
  ::并非所有无限序列都是正值术语序列。 有些包含正值和负值术语 。 请考虑序列 'n=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\在\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Alternating Series
  ::互换序列

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  To this point, previous concepts have identified and discussed properties of infinite series that consist of only non-negative (positive) terms. In this concept, we now expand our coverage to include infinite series that have both positive and negative terms.
  ::至此,以前的概念已经确定并讨论了无限系列的属性,这些属性仅由非负(正)术语组成。 在这一概念中,我们现在扩大了覆盖面,将具有正和负两个术语的无限系列包括在内。

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  Of the many infinite series with both positive and negative terms, the is the simplest type to recognize and study. An a lternating series  is a series whose terms alternate between positive and negative signs.
  ::在许多有正反两词的无限系列中,这是一个最简单的识别和研究类型。 一个交替的系列是一个在正反两词之间交替的系列。

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  An alternating series is an infinite series that can be written as:
  ::交替序列是一个无限的序列,可以写成如下:

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  1. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} u_k = u_1 - u_2 + u_3 - \cdots + (-1)^{k-1} u_k + \cdots\end{align*}$ with $\begin{align*}u_k > 0\end{align*}$ for all $\begin{align*}k\end{align*}$ , or
     ::k=1(- 1) (- 1) k- 1uk=u1- u2+u3} (- 1) k- 1uk+1) , 全部 k 的 uk > 0, 或者
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  2. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k} u_k = -u_1 + u_2 - u_3 + \cdots + (-1)^{k} u_k + \cdots\end{align*}$ with $\begin{align*}u_k > 0\end{align*}$ for all $\begin{align*}k\end{align*}$ .
     ::*k=1(- 1) kukuku1+u2-u3(- 1) kukuk,所有 k 的 uk>0 。

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  Note that the negative sign can be associated with either odd values of the index $\begin{align*}k\end{align*}$ , or even values of the index $\begin{align*}k\end{align*}$ . Also note that the actual numbers represented by the $\begin{align*}u_k\end{align*}$ ’s are positive.
  ::请注意,负号可与指数k的奇值连在一起,甚至与指数k的奇值连在一起。 另请注意,以uk ' 表示的实际数字是正数。

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  There are several types of alternating series whose form should be familiar:
  ::交替系列有几种形式,形式应该熟悉:

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  • alternating harmonic series
    ::交替调和序列

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  Example: $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{1}{k}=1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\end{align*}$
  ::示例:k=1(-1-k-11k=1-12+13-14)

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  This series has terms that look like the harmonic series but the terms with even indices have a negative sign.
  ::这个系列的术语看上去像声波序列, 但单数指数的术语有负信号。

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  • alternating geometric series
    ::交替几何序列

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  Example: $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k} \left(\frac{2}{3}\right)^{k-1}=-1 +\frac{2}{3} - \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 - \cdots\end{align*}$
  ::示例:k=1(- 1)k( 23k- 11+23- (23)2+(23)2+(23)2

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  The odd-indexed terms of this series have the negative sign.
  ::这个系列的奇特索引术语有负信号。

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  • alternating $\begin{align*}p\end{align*}$ - series
    ::交替式p系列

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  Example: $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{1}{\sqrt[3]{k}}=1 -\frac{1}{\sqrt[3]{2}} +\frac{1}{\sqrt[3]{3}} -\frac{1}{\sqrt[3]{4}} - \cdots\end{align*}$
  ::示例:k=1(--1)-11k3=1-123+133-143

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  As with the previous series we have looked at, the important question is how to determine the convergence or divergence of an alternating series. As you might expect, there is an alternating series test.
  ::正如我们所看到的上一个系列一样,重要的问题是如何确定交替系列的趋同或分歧。 正如你可能预期的那样,有一个交替系列的测试。

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  The Alternating Se ries Test  is as follows:
  ::交替序列试验如下:

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  Let $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} u_k\end{align*}$  or  $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k} u_k\end{align*}$ be an alternating series (with $\begin{align*}u_k > 0\end{align*}$  for all $\begin{align*}k\end{align*}$ ), then the series converges if:
  ::Let 'k=1(- 1)-\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\该序列交替序列( \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\该串串串串串串串串会序列的\\\\\\\\\\\\\\\\\\\rererererererererererererererererererererererererererererererererererererererererererere

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  1. $\begin{align*}u_1 \ge u_2 \ge u_3 \ge \ldots \ge u_k \ge \ldots ;\end{align*}$ and
     ::23... 223... 2...
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  2. $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\end{align*}$
     ::立方公尺=0

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  Take the terms of the series and drop their signs. Then the theorem tells us that the terms of the series must be non increasing and the limit of the terms is 0 in order for the test to work. Here is an example of how to use the Alternating Series Test:
  ::接受系列的条款, 并放下它们的标志。 然后理论告诉我们, 系列的条款必须不增加, 并且条件的限度是 0 才能让测试起作用。 下面是如何使用交替序列测试的示例 :

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  Determine if $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k+1} \frac{k+5}{k^3+k}\end{align*}$  converges or diverges.
  ::确定 & k=1 {( - 1) 1k+1k+5k3+k 是否趋同或分歧 。

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  The series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k+1} \frac{k+5}{k^3+k}\end{align*}$  is an alternating series.
  ::& k=1 {( - 1) 1k+1k+5k3+k 序列是一个交替序列 。

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  First we check that the terms of the series are non-increasing. Note that in order for $\begin{align*}u_k \ge u_{k+1}\end{align*}$ , then  $\begin{align*}1 \ge \frac{u_{k+1}}{u_k}\end{align*}$ , or  $\begin{align*}\frac{u_{k+1}}{u_k} \le 1\end{align*}$ .
  ::首先,我们检查该系列的条款是否没有增加。 请注意, 为 uk uk+1, 然后为 uk+1, 或 uk+1uk+1, 或 uk+1uk+1, 请注意 。

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  $$\begin{align*}\frac{u_{k+1}}{u_k}=\frac{\frac{(k+1)+5}{(k+1)^3+(k+1)}}{\frac{k+5}{k^3+k}} = \frac{(k+1)+5}{(k+1)^3+(k+1)} \times \frac{k^3+k}{k+5}\end{align*}$$
  ::uk+1uk=(k+1)+5(k+1)+3(k+13)+(k+1)+(k+1)+5k3+K=(k+1)+5(k+1)+3+(k+1)+(k+1)+3(k3+k+5)

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  Expanding the last expression, we get:
  ::扩展最后一个表达式时, 我们得到 :

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  $$\begin{align*}\frac{u_{k+1}}{u_k}=\frac{(k+6)(k^3+k)}{(k^3+3k^2+4k+2)(k+5)} = \frac{k^4+6k^3+k^2+6k}{k^4+8k^3+19k^2+22k+10}\end{align*}$$
  ::uk+1uk= (k+6)(k3+6)(k3+k3+k)(k3+3k2+4k+2)(k+5)=k4+6k3+k2+K2+6k4+8k3+19k2+22k+10)

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  Since $\begin{align*}k\end{align*}$  is positive and the sum in the numerator is part of the denominator’s sum, the numerator is less than the denominator. Therefore $\begin{align*}\frac{u_{k+1}}{u_k} \le 1\end{align*}$ . Thus, $\begin{align*}u_k > u_{k+1}\end{align*}$ .
  ::k 是正数, 分子中的总和是分母总和的一部分, 分子小于分母。 因此, uk+1uk+1。 因此, uk>uk+1。 因此, uk>uk+1 。

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  Next, we check that $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\end{align*}$ .
  ::下一位,我们检查一下

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  $$\begin{align*}\lim\limits_{k \to +\infty}u_k=\lim\limits_{k \to +\infty}\frac{k+5}{k^3+k}=\lim\limits_{k \to +\infty}\frac{k}{k^3} \frac{1+\frac{5}{k}}{1+ \frac{1}{k^2}}=0.\end{align*}$$
  ::立方公尺=limk@k+5k3+k=limk@kk31+5k1+1k2=0。

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  Another approach to evaluating the limit is to see that $\begin{align*}\lim\limits_{k \to +\infty}u_k=\lim\limits_{k \to +\infty}\frac{k+5}{k^3+k}= \frac{\infty}{\infty}\end{align*}$ , and since this is an indeterminate form, we can use l’Hopital’s rule:
  ::评估限制的另一种方法就是看到Limkáuk=limkäk+5k3+k, 由于这是一种不确定的形式, 我们可以使用医院的规则:

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  $$\begin{align*}\lim\limits_{k \to +\infty}\frac{k+5}{k^3+k}=\lim\limits_{k \to +\infty}\frac{\frac{d}{dk}(k+5)}{\frac{d}{dk}(k^3+k)}=\lim_{k \to +\infty} \frac{1}{3k^2+1}=0.\end{align*}$$
  ::k+5k3+k=limk*ddk(k+5ddk(k3+k)=limk=13k2+1=0)

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  Therefore, by the Alternating Series Test, the alternating series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k+1} \frac{k+5}{k^3+k}\end{align*}$ converges.
  ::因此,在交替序列测试中,交替序列Qk=1(-1)k+1k+5k3+k)相交。

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  Keep in mind that both Alternating Series Test conditions have to be satisfied for the test to prove convergence.
  ::铭记必须满足两个不同的系列试验条件才能证明试验趋同。

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  Take the series   $\begin{align*}\sum\limits_{n=1}^\infty (-1)^{n} \frac{\sqrt{n}}{1+4 \sqrt{n}}\end{align*}$  . We want to determine if it converges or diverges.
  ::取下序列 'n=1\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\ n1+4n。 我们想确定它是否趋同或不同 。

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  The series $\begin{align*}\sum\limits_{n=1}^\infty (-1)^{n} \frac{\sqrt{n}}{1+4 \sqrt{n}}\end{align*}$ is an alternating series.
  ::序列=1(- 1)nn1+4n是一个交替序列。

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  First we check whether the terms of the series are non-increasing. We look at whether $\begin{align*}\frac{u_{k+1}}{u_k} \le 1\end{align*}$ .
  ::首先,我们检查系列条款是否没有增加。我们看看 uk+1uk+1。

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  $$\begin{align*}\frac{u_{n+1}}{u_n}=\frac{\frac{\sqrt{n+1}}{1+4 \sqrt{n+1}}}{\frac{\sqrt{n}}{1+4\sqrt{n}}}=\frac{1+4 \sqrt{n}}{1+4 \sqrt{n+1}} \times \frac{\sqrt{n+1}}{\sqrt{n}}\end{align*}$$
  ::un+1un=n+11+4n+1n1+4n=1+4n1+4n+4n+4n+1xn+1n

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  Expanding the last expression, we get:
  ::扩展最后一个表达式时, 我们得到 :

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  $$\begin{align*}\frac{u_{n+1}}{u_n}&=\frac{1+4 \sqrt{n}}{1+4 \sqrt{n+1}} \times \frac{\sqrt{n+1}}{\sqrt{n}} \\ &= \frac{\frac{1}{\sqrt{n}}+4}{\frac{1}{\sqrt{n+1}}+4} \times \frac{\sqrt{n}}{\sqrt{n+1}} \times \frac{\sqrt{n+1}}{\sqrt{n}} \\ &= \frac{4+ \frac{1}{\sqrt{n}}}{4+ \frac{1}{\sqrt{n+1}}} > 1\end{align*}$$
  ::un+1un=1+4n1+4n1n+1xn+1n+1n=1n+41n+1+4xnn+1xn+1n+1n=4+1n4+1n4+1n+1n+1n+1n+1>1

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  Notice that $\begin{align*}\frac{1}{\sqrt{n}} > \frac{1}{\sqrt{n+1}}\end{align*}$ , which means $\begin{align*}\frac{u_{n+1}}{u_n} > 1\end{align*}$ , which does not satisfy the Alternating Series condition.
  ::注意 1n>1n+1, 这意味着 un+1un>1, 不符合交替序列条件 。

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  Therefore the alternating series  $\begin{align*}\sum\limits_{n=1}^\infty(-1)^n \frac{\sqrt{n}}{1+4 \sqrt{n}}\end{align*}$ diverges.
  ::因此交替序列 'n=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\4nn差异。

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  Notice also that for the second condition of the Alternating Series Test we have
  ::请注意,对于交替系列试验的第二个条件

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  $$\begin{align*}\lim_{n \to + \infty} u_n=\lim_{n \to + \infty} \frac{\sqrt{n}}{1+4 \sqrt{n}}=\frac{\infty}{\infty}\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}Limn_un=limn_n1+4n}Limn_un=limn_n_n1+4n_

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  Since this is an indeterminate form, we can use l’Hopital’s rule:
  ::因为这是一种不确定的形式, 我们可以使用医院的规则:

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  $$\begin{align*}\lim\limits_{n \to +\infty}u_n = \lim_{n \to +\infty}\frac{\sqrt{n}}{1+4 \sqrt{n}} = \lim_{k \to +\infty} \frac{\frac{1}{2}n^{-\frac{1}{2}}}{2n^{-\frac{1}{2}}}=\frac{1}{4} \ne 0\end{align*}$$
  ::======================================================================================================================================================================================================================================= ===============================================================================================================================================================================================================================

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  The second condition also is not satisfied.
  ::第二个条件也不符合。

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  Alternating Series Truncation Error (or Remainder)
  ::转划错误 (或留置器)

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  A partial sum $\begin{align*}s_n\end{align*}$  of an alternating series can be used to approximate the sum of the series. If the alternating series converges, we can actually find a bound on the difference between the partial sum and the actual sum. This difference, or remainder, is called the error.
  ::交替序列的一部分总和可以用来估计序列的总和。如果交替序列相交,我们实际上可以找到部分总和和实际总和之间的差数。这一差数或余数称为错误。

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  Let $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1}u_k\end{align*}$ or $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k}u_k\end{align*}$ be an alternating series ( $\begin{align*}u_k > 0\end{align*}$ for all $\begin{align*}k\end{align*}$ ) that satisfies the conditions of the Alternating Series Test.
  ::Let 'k=1(- 1)-\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\Kukuk 交替序列(uk>0所有 k)满足交替序列测试的条件。

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  If the series has sum $\begin{align*}S\end{align*}$ , and the $\begin{align*}n\end{align*}$ th partial sum is $\begin{align*}s_n\end{align*}$ , then
  ::如果该序列有S和S,且n为Sn,那么,如果该序列有S和S,且n部分为Sn,则

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  $$\begin{align*}|S-s_n| \le u_{n+1}.\end{align*}$$
  ::S-snun+1。

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  The main idea of the theorem is that the remainder $\begin{align*}|S-s_n|\end{align*}$  cannot get larger than the next, $\begin{align*}n+1\end{align*}$ st, term in the series.
  ::理论的主要理念是,其余的 +S-sn不能大于 系列中的下一个 n+1 任期。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked what type of series the  series $\begin{align*}\sum\limits_{n=1}^\infty 8 \left(-\frac{1}{2}\right)^{n-1}\end{align*}$  is and how to determine if it converges or diverges. 
  ::更早之前,有人问您系列"n=1"=1"8(-12)-n-1是哪一类系列以及如何确定它是否趋同或不同。

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  Did you recognize this series as a geometric series? It is an alternating geometric series, where the terms alternate between positive and negative. Like any geometric series, convergence/divergence is determined by the common ratio, $\begin{align*}r:|r|<1 \Rightarrow \end{align*}$ convergence;
  ::您是否承认这个序列是一个几何序列? 这是一个交替的几何序列, 其术语在正数和负数之间交替。 与任何几何序列一样, 趋同/ 调和由共同比率( r: r: r: 1 ) 趋同决定 ;

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  $\begin{align*}|r| \ge 1 \Rightarrow \end{align*}$ divergence. The series $\begin{align*}\sum\limits_{n=1}^\infty 8 \left(-\frac{1}{2}\right)^{n-1}\end{align*}$ converges because $\begin{align*}r=-\frac{1}{2}\end{align*}$ .
  ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\因为r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\因为r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Example 2
  ::例2

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  For the series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{k+5}{k^3+k}\end{align*}$
  ::对于 & k=1\\\\\\\\\\ 1\1k- 1k+5k3+k 序列来说,\k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  1. Compute the third partial sum, $\begin{align*}s_3\end{align*}$
     ::计算第三份部分总和,S3
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  2. Determine the bound on the remainder.
     ::确定余下部分的约束。

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  We know from earlier in the concept  that the series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{k+5}{k^3+k}\end{align*}$ passes the Alternating Series test and therefore converges.
  ::我们从先前的概念中知道,“k”=1(-1)(-1)k-1k+5k3+k)系列通过交替序列测试,因此会趋同。

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  1. First, we compute the third partial sum to approximate the sum $\begin{align*}S\end{align*}$  of the series:
     ::首先,我们计算第三部分总和,以大致相当于该系列的S总和:

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  $$\begin{align*}s_3 &= (-1)^0 \frac{1+5}{1^3+1} + (-1)^1 \frac{2+5}{2^3+2} + (-1)^2 \frac{3+5}{3^3+3} \\ &= \frac{6}{4} - \frac{7}{10} + \frac{8}{30} \\ &= \frac{16}{15}\end{align*}$$
  ::s3=(-1)01+513+1+1+(-1)12+523+2+(-1)23+533+3=64-710+830=1615

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  2. Next, the theorem tells us to use the $\begin{align*}u_4\end{align*}$  term in the series to calculate the bound on the difference or remainder.
     ::接下来,定理告诉我们使用序列中的 u4 术语来计算差额或剩余部分的界限。

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  Thus $\begin{align*}u_4=\frac{4+5}{4^3+4}=\frac{9}{68}\end{align*}$ .
  ::因此,u4=4+543+4=968。

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  Then $\begin{align*}|S-s_n|=\Big|S-\frac{16}{15}\Big|<\frac{9}{68} \approx 0.13.\end{align*}$
  ::然后是S -S -S -1615 -968 -0.13

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  This tells us that the absolute value of the error or remainder is less than 0.13.
  ::这表明错误或剩余部分的绝对值低于0.13。

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  Example 3
  ::例3

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  For the series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{k^2}{8^k}\end{align*}$ :
  ::对于 & k=1\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1k28KK\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  1. Determine if the series converges or diverges
     ::确定序列相近或相异
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  2. Approximate the sum of the series to four decimal places.
     ::近似于该序列的总和到小数点后四位数。

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  The series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{k^2}{8^k}\end{align*}$  is an alternating series with  $\begin{align*}u_k = \frac{k^2}{8^k}\end{align*}$ .
  ::& k=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\K28KKKKK是一个交替序列 和uk=K28K。

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  First we check that the terms of the series are non-increasing.
  ::首先,我们检查系列条款没有增加。

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  $$\begin{align*}\frac{u_{k+1}}{u_k}= \frac{\frac{(k+1)^2}{8^{k+1}}}{\frac{k^2}{8^k}}= \frac{8^k}{8^{k+1}} \times \frac{(k+1)^2}{k^2} = \frac{1}{8} \left(1+\frac{1}{k}\right)^2<1\end{align*}$$
  Therefore $\begin{align*}\frac{u_{k+1}}{u_k}<1\end{align*}$ . Thus,  $\begin{align*}u_k > u_{k+1}\end{align*}$ .
  ::uk+1uk= (k+1,28k+1k28k=8k8k8k=8k8kx1x(k+1,2k2=18(1+1k)2 <1 因此uk+1uk <1。 因此, uk>uk+1uk+1k+1。 因此, uk>uk+1。

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  Next, we check that $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\end{align*}$ .
  ::下一位,我们检查一下

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  $$\begin{align*}\lim\limits_{k \to +\infty}u_k=\lim\limits_{k \to +\infty} \frac{k^2}{8^k}=\frac{\infty}{\infty}\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  Since this is an indeterminate form, we can use l’Hopital’s rule:
  ::因为这是一种不确定的形式, 我们可以使用医院的规则:

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  $$\begin{align*}\lim\limits_{k \to +\infty} \frac{k^2}{8^k}=\lim\limits_{k \to +\infty} \frac{\frac{d}{dk}(k^2)}{\frac{d}{dk}(8^k)} = \lim\limits_{k \to +\infty} \frac{2k}{8^k \ln 8} = \lim\limits_{k \to +\infty} \frac{2}{8^k(\ln 8)^2}=0\end{align*}$$
  :k2)ddk(8k) =limk=2k8kln(8k) =8=limk=28k(ln_8)2=0

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  By the Alternating Series Test, the alternating series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k-1}\frac{k^2}{8^k}\end{align*}$ converges.
  ::在交替序列测试中,交替序列 'k=1\\\\\\\\\\\\\\\1\\\1\K-1k28k会合。

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  To approximate the sum of the series to four decimal places means, $\begin{align*}|S-s_n| \le u_{n+1} \le 0.00005\end{align*}$ . We need to find the series term $\begin{align*}n\end{align*}$  such that $\begin{align*}\frac{n^2}{8^n} \le 0.00005\end{align*}$ . If $\begin{align*}n=7\end{align*}$ then $\begin{align*}\frac{n^2}{8^n} = 0.00002 < 0.00005\end{align*}$ .
  ::约等于小数点后四位数的数组之和,是指 +S-snun+10.005。我们需要找到序列值为 n28n0.005。如果 n=7,那么 n28n=0.0002 < 0.0005,那么n=7, n28n=0.002<0.0005。

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  Therefore 6 terms of the series are needed: $\begin{align*}\sum\limits_{k=1}^6(-1)^{k-1} \frac{k^2}{8^k}\end{align*}$  is accurate to 4 decimal places.
  ::因此,该系列需要6个条件: k=16(-1)k-1k28k 精确到小数点后4位。

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  Adding the first 6 terms: $\begin{align*}\sum\limits_{k=1}^6(-1)^{k-1} \frac{k^2}{8^k}=0.0768\end{align*}$
  ::添加前 6 个术语 : k=16 (- 1) 1k- 1k28k=0.0 768

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  Review
  ::回顾

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  For #1-11, determine if the series converges or diverges using the Alternating Series Test.
  ::对于 #1-11, 使用交替序列测试来确定该序列是否趋同或不同 。

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  1. $\begin{align*}\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k}\end{align*}$
     ::*k=1(- 1)k+1k
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  2. $\begin{align*}\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{2k+1}\end{align*}$
     ::*k=1(- 1) k+12k+1
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  3. $\begin{align*}\sum\limits_{n=1}^\infty (-1)^n \frac{n}{n+3}\end{align*}$
     ::Nn=1(- 1) nn+3
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  4. $\begin{align*}\sum\limits_{k=1}^\infty \left(-\frac{5}{13}\right)^{k-1}\end{align*}$
     ::k=1(- 513)k- 1
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  5. $\begin{align*}\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n^6}\end{align*}$
     ::n=1(- 1)n+1n6
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  6. $\begin{align*}\sum\limits_{k=1}^\infty \frac{(-1)^{k}}{10^n n!}\end{align*}$
     ::K=1(- 1)k10nn!
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  7. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^k \frac{1}{\ln(n+1)}\end{align*}$
     ::*k=1(- 1)k1ln( n+1)
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  8. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^k \frac{\ln(n+1)}{n+1}\end{align*}$
     ::*k=1(- 1)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ n+1\n+1n+1n+1)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  9. $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{n} \cos n \pi\end{align*}$
     ::=1 =1 =1 ncos = = = = = = = = = 1ncos = = = = = = = = = = = = = = = = = = = = = 1ncos = = = = = = = = = = = = = = = = =
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  10. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k+1} \frac{1}{n \sqrt{n}}\end{align*}$
      ::*k=1(- 1) k+11nn
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  11. $\begin{align*}\sum\limits_{k=1}^\infty (-1)^{k+1} \frac{3n+2}{n+10}\end{align*}$
      ::*k=1(- 1) k+13n+2n+10
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  12. Show that the series $\begin{align*}\sum\limits_{k=1}^\infty (-1)^k \frac{5}{k^2}\end{align*}$ converges according to the Alternating Series Test. Let $\begin{align*}S=\sum\limits_{k=1}^\infty (-1)^k \frac{5}{k^2}\end{align*}$ . Compute $\begin{align*}s_3\end{align*}$ for $\begin{align*}\sum\limits_{k=1}^\infty (-1)^k \frac{5}{k^2}\end{align*}$ and determine the bound on $\begin{align*}|S-s_3|\end{align*}$ .
      ::显示 & k=1\\\\\\\\\\\\\\\\\\\\\1K5K2 序列会根据交替序列测试对齐 。 Let Säk=1\\\\\\\\\\\\\\\\\\1\\\\\\\1\\\\\\\\\\1\K5K2 计算 s3 来计算\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  13. Show that the series $\begin{align*}\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k !}\end{align*}$ converges according to the Alternating Series Test. Let $\begin{align*}S=\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k !}\end{align*}$ . Compute $\begin{align*}s_4\end{align*}$ for  $\begin{align*}\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k !}\end{align*}$ and determine the bound on $\begin{align*}|S-s_4|\end{align*}$ .
      ::显示 & k= 1\\\\\\\\\\\\\\1\\\\\\\\\\\\1K+1k! 根据交替序列测试, 序列 & k=1\\\\\\\\\\\\\\\\\1K+1k! 集合 。 Let Säk=1\\\\\\\\\\\\\\\\\\\\\\\1K+1k! 计算 s4 =\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\- S\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  For the series in #1, find the least value of $\begin{align*}n\end{align*}$  such that:
  ::对于 # 1 中的序列, 找到n 的最低值, 如下 :

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  14. $\begin{align*}\Bigg| \sum\limits_{k=1}^n \frac{(-1)^{k+1}}{k} -S \Bigg| <0.05\end{align*}$
      ::k=1n(- 1) +1k- S @ @ 0.0 5)
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  15. $\begin{align*}\Bigg| \sum\limits_{k=1}^n \frac{(-1)^{k+1}}{k} -S \Bigg| <0.005\end{align*}$
      ::@k=1n(- 1) +1k- S @ @ @ 0.0005)
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  16. $\begin{align*}\Bigg| \sum\limits_{k=1}^n \frac{(-1)^{k+1}}{k} -S \Bigg| <0.0001\end{align*}$
      ::k=1n(- 1) +1k- S @ 0. 001)

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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 。