9.5 无限系列的某些属性

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  无限系列的某些属性

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  The previous concept established the basis for determining whether an infinite series converges or diverges . Well, if the convergence or divergence of some infinite series has been determined, are there basic properties that tell us how they might be combined? Recall that the same question occurs for how functions, limits, and integrals can be combined. Here, think in terms of how to combine series that are convergent and/or divergent series. If two convergent series are combined, is the resulting series convergent? What about combining a convergent and a divergent series?
  ::先前的概念为确定一个无限序列的趋同或相异奠定了基础。 如果某些无限序列的趋同或差异已经确定, 那么, 如果某些无限序列的趋同或差异已经确定, 是否有基本属性可以告诉我们如何将它们结合起来? 提醒大家注意, 如何将函数、 限制和整体体结合起来, 也会出现同样的问题 。 这里, 思考如何将集合和/ 或不同序列的序列结合起来。 如果将两个相趋同的序列合并, 由此产生的序列是不是趋同的? 如何合并一个趋同和不同的序列呢 ?

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  Properties of Infinite Series
  ::无限系列的属性

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  When working with infinite series, the following basic properties are useful to know:
  ::在使用无限序列时,以下基本特性有助于了解:

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  • $\begin{align*}n\end{align*}$ th- Term Property of a Convergent Infinite Series
    ::N至n 一致的无限制丛书的定期财产
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  • Basic Summation/Product Properties of Infinite Series
    ::无限系列基本总和/生产属性
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  • Reindexing an Infinite Series
    ::重新编制无限系列索引

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  $\begin{align*}n\end{align*}$ th-Term Property of a Convergent Infinite Series
  ::N至n 一致的无限制丛书的定期财产

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  In the previous concept, a test for the divergence of an infinite series was presented called the $\begin{align*}n\end{align*}$ th Term Test for Divergence . This test is actually a result of the following property of a convergent infinite series:
  ::在前一个概念中,对无限序列的差异进行了测试,称为 nth Term 差异测试。这个测试实际上是聚合无限序列的以下属性的结果:

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  If the infinite series  $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ converges, then $\begin{align*}\lim\limits_{k\rightarrow +\infty} u_k=0\end{align*}$ .
  ::如果无限序列@k=1uk 集合, 那么limk@uk=0 。

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  Note: $\begin{align*}\lim\limits_{k\rightarrow +\infty}u_k=0\end{align*}$  is not a test for series convergence, only a property of a convergent series. A series may diverge but still have $\begin{align*}\lim\limits_{k\rightarrow +\infty}u_k=0\end{align*}$ .
  ::注: limkuk=0 不是序列趋同的测试, 只是一个集合序列的属性。 一个序列可能会出现差异, 但仍然有 limkuk=0 。

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  As noted, we cannot use this property of a convergent series as a test for convergence, only a corroboration of convergence once it has been determined by using an appropriate test.
  ::如前所述,我们不能将这种趋同系列的特性作为趋同的检验标准,而只有在使用适当试验确定趋同之后,才能证实趋同。

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  Let's show that the convergent $\begin{align*}p\end{align*}$ -series  $\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{k^3}\end{align*}$ satisfies the  $\begin{align*}n\end{align*}$ th term property of a convergent series.
  ::让我们展示一下 集合的 p系列 p-series \\ k= 11k3 符合集合序列的 nth 术语属性 。

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  The $\begin{align*}n\end{align*}$ th term of the convergent series $\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{k^3}\end{align*}$  is given by $\begin{align*}u_k=\frac{1}{k^3}\end{align*}$ , and $\begin{align*}\lim\limits_{k\rightarrow +\infty}u_k=\lim\limits_{k\rightarrow +\infty}\frac{1}{k^3}=0\end{align*}$ .
  ::以 uk= 1k3 和limk@uk=limk*1k3=0 给定了合并序列的下一个期限。

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  Therefore, the convergent series  $\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{k^3}\end{align*}$ satisfies the  $\begin{align*}n\end{align*}$ th term property of a convergent series
  ::因此, 集合序列 *k=11k3sasfaisfies 一个集合序列的 nth 术语属性

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  Now, let's show that the divergent harmonic-series  $\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{k}\end{align*}$ also satisfies the $\begin{align*}n\end{align*}$ th term property of a convergent series.
  ::现在,让我们展示一下 不同的调和序列k=11k 也满足了集合序列的 nth 术语属性。

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  The  $\begin{align*}n\end{align*}$ th term of the divergent series $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{k}\end{align*}$  is given by $\begin{align*}u_k=\frac{1}{k}\end{align*}$ , and $\begin{align*}\lim\limits_{ k\rightarrow +\infty} u_k =\lim\limits_ {k\rightarrow +\infty} \frac{1}{k^3}=0\end{align*}$ .
  ::不同序列的 nth 期由 uk = 1k 和 limk uk = limk = 1k = 0 给出 。

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  Therefore, the divergent series  $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{k^3}\end{align*}$ also satisfies the  $\begin{align*}n\end{align*}$ th term property of a convergent series.
  ::因此,不同的序列Qk=11k3也满足了集合序列的Nth术语属性。

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  Do not use the  $\begin{align*}n\end{align*}$ th term property of a convergent infinite series as a test for convergence!
  ::不使用集合无限序列的第 n 术语属性作为趋同的测试 !

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  Basic Summation/Product Properties of Infinite Series
  ::无限系列基本总和/生产属性

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  As with sequences, there are some useful summation and product properties that apply to infinite series.
  ::与序列一样,有些有用的总和和和产品特性适用于无限序列。

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  The sum or difference of convergent series is also convergent:
  ::聚合序列的总和或差数也是聚合的:

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  • Suppose $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$  are convergent series with $\begin{align*}\sum\limits _{k=1}^\infty u_k=S_u\end{align*}$  and $\begin{align*}\sum\limits _{k=1}^\infty v_k=S_v\end{align*}$ , then:
    ::假设k=1uk和k=1vk为集合序列,与k=1uk=Su和k=1vk=Sv合并,然后:

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  $\begin{align*}\sum\limits_{k=1}^\infty (u_k+v_k)\end{align*}$  is convergent and $\begin{align*}\sum\limits _{k=1}^\infty(u_k+v_k)=S_u+S_v\end{align*}$
  ::*\ k= 1( uk+vk) 是集合值, @ k= 1( uk+vk) = Su+Sv

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  $\begin{align*}\sum\limits_{k=1}^\infty(u_k-v_k)\end{align*}$  is convergent and $\begin{align*}\sum\limits_{k=1}^\infty(u_k-v_k)=S_u-S_v\end{align*}$
  ::*\ k= 1( uk- vk) 是集合, 和 *k= 1( uk- vk) = Su- Sv

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  • The sum of a convergent and divergent series is divergent.
    ::趋同和不同系列的总和是不同的。

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  Suppose $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  is convergent and $\begin{align*}\sum\limits_{k=1}^\infty v_k\end{align*}$  is divergent, then:
  ::假设“k=1uk”是趋同的,“k=1vk”是不同的,那么:

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  $\begin{align*}\sum\limits_{k=1}^\infty(u_k+v_k)\end{align*}$  is divergent.
  ::k=1( uk+vk) 存在差异 。

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  Note that the sum or difference of two divergent series may or may not be divergent.
  ::请注意,两个不同系列的总和或差额可能不同,也可能不同,也可能不同。

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  • Multiplying a series by a nonzero constant does not affect convergence or divergence:
    ::用非零常数乘以一个序列并不影响趋同或分歧:

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  Let $\begin{align*}c\ne 0\end{align*}$  be a constant, then:
  ::让我们成为常数,然后:

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  If $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  converges and $\begin{align*}\sum\limits_{k=1}^\infty u_k = S_u\end{align*}$ , then $\begin{align*}\sum\limits_{k=1}^\infty {cu}_k\end{align*}$  converges and $\begin{align*}\sum\limits_{k=1}^\infty {cu}_k = cS_u\end{align*}$ .
  ::如果“k”=1uk 集合和“k”=1uk=Su,那么“k”=1cuk 集合和“k”=1cuk=cSu。

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  If $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$  diverges, then $\begin{align*}\sum\limits_{k=1}^\infty cu_k\end{align*}$  also diverges.
  ::如果“k”=1uk的差异,那么“k”=1cuk的差异也不同。

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  • Adding or subtracting a finite number of terms from an infinite series does not affect convergence or divergence.
    ::从无限系列中增加或减去一定数目的术语并不影响趋同或分歧。

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  If  $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ converges, then $\begin{align*}\sum\limits_{k=1}^\infty u_k \pm(u_1+u_2+\ldots+u_m)\end{align*}$  is also convergent.
  ::如果 & k= 1uk 相交, 那么 & k= 1uk( u1+u2+...+...+um) 也是会合 。

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  If  $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ diverges, then $\begin{align*}\sum\limits_{k=1}^\infty u_k \pm(u_1+u_2+\ldots+u_m)\end{align*}$  is also divergent.
  ::@k=1uk 差异, 那么@k=1uk( u1+u2+...+...+um) 也有所不同 。

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  Note that adding or removing a finite number of terms will affect the sum.
  ::请注意,增加或删除若干限定任期将影响总和。

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  Applying the properties above, let's find the sum of $\begin{align*}\sum\limits_{k=1}^\infty\left(\frac{2}{3^{k-1}}+\frac{1}{8^{k-1}}\right)\end{align*}$  if it exists.
  ::应用上面的属性时, 如果存在的话, 我们来查找%k=1( 23k- 1+18k- 1) 的和 。

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  Examining the two series we see that
  ::检查我们所看到的两个系列

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  $\begin{align*}\sum\limits_{k=1}^\infty \frac{2}{3^{k-1}}\end{align*}$  is a convergent geometric series with  $\begin{align*}a=2\end{align*}$ and $\begin{align*}r=\frac{1}{3}\end{align*}$ , and sum is $\begin{align*}\frac{a}{1-r}=\frac{2}{1-\frac{1}{3}}=3\end{align*}$ .
  ::k=123k- 1 是集合几何序列, a=2 和 r= 13, 和 a1-r=21- 13=3。

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  $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{8^{k-1}}\end{align*}$  is a convergent geometric series with  $\begin{align*}a=1\end{align*}$ and $\begin{align*}r=\frac{1}{8}\end{align*}$ , and sum is $\begin{align*}\frac{a}{1-r}=\frac{1}{1-\frac{1}{8}}=\frac{8}{7}\end{align*}$ .
  ::k=118k- 1 是一个集合几何序列, a= 1 和 r= 18, 和 a1-r= 11-18= 87。

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  Therefore, using the Basic Properties Theorem, $\begin{align*}\sum\limits_{k=1}^\infty \left(\frac{2}{3^{k-1}}+\frac{1}{8^{k-1}}\right)=\sum\limits_{k=1}^\infty \frac{2}{3^{k-1}}+\sum\limits_{k=1}^\infty \frac{1}{8^{k-1}}=3+\frac{8}{7}=\frac{29}{7}\end{align*}$ .
  ::因此,使用基本属性理论, @k=1(23k-1+18k-1) @k=123k-1Qk=118k-1=3+87=297。

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  Now, let's show that subtracting terms $\begin{align*}k=11,12,13,\ldots ,19\end{align*}$  from the series $\begin{align*}\sum\limits_{k=1}^\infty \frac{3}{5^{k-1}}\end{align*}$  does not affect the convergence of this geometric series.
  ::现在,让我们显示,从序列中减去 k=11, 12, 13,...,..., 19 并不影响这个几何序列的趋同 。

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  $\begin{align*}\sum\limits_{k=1}^\infty \frac{3}{5^{k-1}} = \sum\limits_{k=1}^\infty 3 \left(\frac{1}{5}\right)^{k-1}\end{align*}$ is a geometric series with $\begin{align*}a=3\end{align*}$ , $\begin{align*}r=\frac{1}{5}\end{align*}$ , and sum $\begin{align*}\frac{a}{1-r}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}\end{align*}$ . If terms
  ::k=135k- 1Qk=13( 15k- 1) 是一个几何序列, a= 3, r= 15, sum a1- r=31-15=154。 如果条件如此, 则 a1- r=31- 15=154 。

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  $\begin{align*}k=11,12,13,\ldots ,19\end{align*}$  are removed from the series, the new series becomes $\begin{align*}\sum\limits_{k=1}^{10} 3\left(\frac{1}{5}\right)^{k-1} + \sum\limits_{k=20}^\infty 3\left(\frac{1}{5}\right)^{k-1}\end{align*}$ .
  ::k=11, 12, 13,...,...,...,..., 19 从序列中删除, 新的序列变成 10 k=13( 15k- 1QQ=203( 15k- 1) 。

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  The first series  $\begin{align*}\sum\limits_{k=1}^{10} 3 \left(\frac{1}{5}\right)^{k-1}\end{align*}$ is a finite geometric series with sum $\begin{align*}\frac{a\left(1-r^n\right)}{1-r}=\frac{3\left[1-\left(\frac{1}{5}\right)^{10}\right]}{1-\frac{1}{5}}=\frac{15}{4}\left[1-\left(\frac{1}{5}\right)^{10}\right]\end{align*}$ .
  ::第一个序列 10Qk=13(15k-1)是一个限定的几何序列,A(1-rn)1-r=3[1-(15)10]1-15=154[1-(15)10]。

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  The second series  $\begin{align*}\sum\limits_{k=20}^\infty 3 \left(\frac{1}{5}\right)^{k-1}\end{align*}$ can be written as
  ::第二系列 k=203(15k-1)可以写为

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  $$\begin{align*}\sum\limits_{k=20}^\infty 3\left(\frac{1}{5}\right)^{k-1}&=3\left(\frac{1}{5}\right)^{19}\left[1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\ldots\right]\\ &=\sum\limits_{k=1}^\infty 3\left(\frac{1}{5}\right)^{19}\left(\frac{1}{5}\right)^{k-1}\end{align*}$$
  ::k=203( 15k- 1) 3( 15) 19[ 1+15+( 15) 2+( 153+... ]] @k=13( 15) 19( 15) k- 1

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  But $\begin{align*}\sum\limits_{k=1}^\infty 3\left(\frac{1}{5}\right)^{19}\left(\frac{1}{5}\right)^{k-1}\end{align*}$  is also a convergent geometric series with $\begin{align*}a=3\left(\frac{1}{5}\right)^{19}\end{align*}$ , $\begin{align*}r=\frac{1}{5}\end{align*}$ , and sum $\begin{align*}\frac{a}{1-r}=3\left(\frac{1}{5}\right)^{19}\frac{1}{1-\frac{1}{5}}=\frac{15}{4}\left(\frac{1}{5}\right)^{19}\end{align*}$ .
  ::但k=13(15)19(15)19(15)1-1也是一个集合几何序列,有a=3(15)19,r=15,和a1-r=3(15)1911-15=154(15)19。

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  Therefore the series  $\begin{align*}\sum\limits_{k=1}^{10}3\left(\frac{1}{5}\right)^{k-1} + \sum\limits_{k=20}^\infty 3 \left(\frac{1}{5}\right)^{k-1}\end{align*}$ can be written as
  ::因此, 10Qk=13( 15k- 1 @ @ @ @ @ @% k=203( 15k- 1) 序列可以以下列方式写入 10Qk=13( 15k- 1) :

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  $$\begin{align*}\sum_{k=1}^{10} 3 \left(\frac{1}{5}\right)^{k-1} + \sum_{k=20}^\infty 3 \left(\frac{1}{5}\right)^{k-1} &= \frac{15}{4}\left[1-\left(\frac{1}{5}\right)^{10}\right]+ \frac{15}{4}\left(\frac{1}{5}\right)^{19}\\ &= \frac{15}{4}\left[1-\left(\frac{1}{5}\right)^{10}+ \left(\frac{1}{5}\right)^{19}\right] \ldots \text{This is a finite sum, which show}\\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{series convergence!}\end{align*}$$
  ::10k=13( 15- 1k=203( 15- 1) 15- 1=154[ 1- (15)10]+154( 15)19=154[ 1- (15)10+(1519)]...这是一个有限数额, 显示序列趋同 !

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  Therefore, the convergent series  $\begin{align*}\sum\limits_{k=1}^\infty \frac{3}{5^{k-1}} = \sum\limits_{k=1}^\infty 3 \left(\frac{1}{5}\right)^{k-1}\end{align*}$ with terms $\begin{align*}k=11,12,13,\ldots ,19\end{align*}$  removed is a new series that is also convergent!
  ::因此, 集合序列@k=135k- 1k=13( 15k- 1) 和 k=11, 12, 13,...,..., 19 被删除, 是一个新序列, 同时也是集合 !

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  Reindexing an Infinite Series
  ::重新编制无限系列索引

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  The last property to be addressed in this concept is reindexing of an infinite series. By this we mean changing the index of the series to start at different number than the current series, e.g. $\begin{align*}\sum\limits_{n=1}^\infty a_n = \sum\limits_{n=0}^\infty a_{n+1}\end{align*}$ . As long as we preserve the order of the terms, the index of a series can be changed to start at different number as follows: $\begin{align*}\sum\limits_{n=p}^\infty a_n = \sum\limits_{k=s}^\infty a_{k+(p-s)}\end{align*}$ .
  ::此概念中最后一个要处理的属性是无限序列的重新索引。 我们的意思是将序列的索引从当前序列的不同数字开始, 例如 n=1ann=0an+1. 只要我们保留术语的顺序, 一个序列的索引可以更改为以以下不同数字开始: n=pank=sak+(p-s) 。

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  Now, let's r eindex the series $\begin{align*}\sum\limits_{n=1}^\infty \frac{4}{3^{n-1}}\end{align*}$  so that the index starts at 6 rather than 1.
  ::现在,让我们重新索引 系列 'n=143n -1' 这样指数开始在 6 而不是 1 。

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  We want the first index value to be 6.
  ::我们希望第一个指数值为6。

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  Establish a new index $\begin{align*}k\end{align*}$  so that  $\begin{align*}k=s=6\end{align*}$ corresponds to $\begin{align*}n=p=1\end{align*}$ .
  ::建立新索引 k, 这样 k=s=6 与 n=p=1 相对应 。

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  Then the new series with starting index  $\begin{align*}k=6\end{align*}$ can now be rewritten as:
  ::然后,有起始索引k=6的新序列现在可以重写为:

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  $$\begin{align*}\sum_{n=1}^\infty \frac{4}{3^{n-1}}= \sum_{k=6}^\infty \frac{4}{3^{{k+(1-6)-1}}}= \sum_{k=6}^\infty \frac{4}{3^{k-6}}\end{align*}$$
  ::@n=143n-1k=643k+(1-6)-1k=643k-6

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  You can check that the series on the right is the same series as the one on the left by writing out the first few terms for each series. Notice that the terms are still in order.
  ::您可以通过为每个序列写出前几个术语来检查右侧的序列与左侧的序列相同。 请注意, 术语仍然符合顺序 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked if there are basic properties that tell us how the convergence or divergence of some infinite series can be combined. Recall that the same question occurs for how functions, limits, derivatives and integrals can be combined. If two convergent series are added or subtracted, the resulting series is also convergent. If a convergent series and a divergent series are added together, the resulting series is divergent. The product of a series and a nonzero constant does not affect convergence or divergence. 
  ::早些时候, 有人询问您是否有基本属性可以告诉我们如何将某些无限序列的趋同或分歧合并在一起。 记得对于如何将函数、 限值、 衍生物 和集成物 合并起来, 也会出现同样的问题。 如果增减两个集合序列, 由此产生的序列也是趋同的。 如果将一个趋同序列和不同的序列合并在一起, 由此产生的序列是不同的。 一个序列和一个非零常数的产物不会影响趋同或分歧 。

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  Example 2
  ::例2

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  Find the sum of $\begin{align*}\sum\limits_{k=1}^\infty 2\left(\frac{5}{6^{k-1}}\right)\end{align*}$ .
  ::查找 k=12(56k- 1)的和。

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  By the rules for constant in infinite series, $\begin{align*}\sum\limits_{k=1}^\infty 2 \left(\frac{5}{6^{k-1}}\right) = 2\sum\limits_{k=1}^\infty \frac{5}{6^{k-1}}\end{align*}$ .
  ::依据无穷序列中的恒定值规则, @k=12( 56k- 1)=2@k=156k- 1) 。

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  The series  $\begin{align*}\sum\limits_{k=1}^\infty \left(\frac{5}{6^{k-1}}\right)\end{align*}$ is a geometric series with $\begin{align*}a=5\end{align*}$  and $\begin{align*}r=\frac{1}{6}\end{align*}$ , so that it converges.
  ::=k=1(56k- 1)序列是一个具有 a=5 和 r=16 的几何序列,因此它会汇合。

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  Note that, by the Theorem on convergence of geometric series, $\begin{align*}\sum\limits_{k=1}^\infty \frac{5}{6^{k-1}}=\frac{5}{1-\frac{1}{6}}=\frac{5}{\frac{5}{6}}=6\end{align*}$ .
  ::注意,根据几何序列趋同理论, k=156k-1=51-16=556=6。

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  Then $\begin{align*}\sum\limits_{k=1}^\infty 2 \left(\frac{5}{6^{k-1}}\right)=2\times 6=12\end{align*}$ .
  ::然后是k=12(56k- 1)=2x6=12。

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  Example 3
  ::例3

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  Reindex the series to start at 25.
  ::25时开始的序列重新索引。

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  We want the series $\begin{align*}\sum\limits_{k=1}^\infty 2\left(\frac{5}{6^{k-1}}\right)\end{align*}$  to have a starting index of 25.
  ::我们希望 & k=12( 56k- 1) 序列的起始指数为 25 。

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  If the new index is $\begin{align*}n\end{align*}$ , then the starting index is $\begin{align*}n=25\end{align*}$ .
  ::如果新指数为 n,则起始指数为 n=25。

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  Then the new series with starting index  $\begin{align*}n=25\end{align*}$ can be rewritten as:
  ::然后,带有起始指数n=25的新序列可以重写为:

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  $\begin{align*}\sum\limits_{k=1}^\infty 2 \left(\frac{5}{6^{k-1}}\right) = \sum\limits_{n=25}^\infty 2 \left(\frac{5}{6^{n+(1-25)-1}}\right)=\sum\limits_{n=25}^\infty 2 \left(\frac{5}{6^{n-25}}\right)\end{align*}$
  ::k=12(56k- 1) n=252(56n+(1-25)-1) n=252(56n- 25)

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  Always check to make sure that the series on the right is the same series as the one on the left by writing out the first few terms for each series. The terms must still be in order.
  ::总是检查以确保右侧的序列与左侧的序列相同, 写出每个序列的最初几个术语。 这些术语必须保持顺序 。

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  Review
  ::回顾

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  For #1-9, does the series converge or diverge? If a series converges, find its sum.
  ::对于 #1-9, 该序列是趋同还是相左? 如果一个序列相趋同, 就会找到它的总和 。

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  1. $\begin{align*}\sum\limits_{n=1}^\infty\frac{3n}{n+4}\end{align*}$
     ::*n=13nn+4 *n=13nn+4
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  2. $\begin{align*}\sum\limits_{n=1}^\infty\frac{2+(-4)^n}{5^n}\end{align*}$
     ::*n=12+(-4)n5n
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  3. $\begin{align*}\sum\limits_{n=1}^\infty\frac{n}{4}\end{align*}$
     ::#n=1n4 # #n=1n4 #n=1n4 #n=1n4 #n=1n4 #n=1n4 #n=1n4 #n=1n4 #n=1n4
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  4. $\begin{align*}\sum\limits_{n=1}^\infty\frac{3n}{n^4}\end{align*}$
     ::*n=13nn4 *n=13nn4 *n=13nn4 *n=13nn4 *n=13nn4 *n=13nn4 *n=13nn4 *n=13nn4
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  5. $\begin{align*}\sum\limits_{n=1}^\infty\left[\frac{n^2+2}{n^2+1}+5\left(\frac{1}{2}\right)^{n-1}\right]\end{align*}$
     ::n=1[n2+2n2+1+5(12)-1]
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  6. $\begin{align*}\sum\limits_{n=1}^\infty\frac{n^2+3n+2}{4n^2+1}\end{align*}$
     ::*n=1n2+3n+24n2+1
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  7. $\begin{align*}\sum\limits_{n=1}^\infty\left[\frac{2}{n^2}+7\left(\frac{1}{3}\right)^{n+3}\right]\end{align*}$
     ::n=1[2n2+7(13n+3)
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  8. $\begin{align*}\sum\limits_{k=2}^\infty\left(\left(-\frac{2}{3}\right)^{k-1}+\frac{1}{5^{k-1}}\right)\end{align*}$
     ::*k=2((- 23k- 1+15k- 1)
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  9. $\begin{align*}\sum\limits_{k=1}^\infty\left(\frac{4}{5^{k-1}}-\frac{2^k}{3^k}\right)\end{align*}$
     ::*k=1(45k-1-2k3k)
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  10. Use the fact that $\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}\end{align*}$  to find $\begin{align*}\sum\limits_{n=3}^\infty\frac{4}{n^2}\end{align*}$ .
      ::使用 {n=11n2}\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  11. Reindex $\begin{align*}\sum\limits_{n=1}^\infty\frac{3n}{n+4}\end{align*}$  to be a sum of the form $\begin{align*}\sum\limits_{n=7}^\infty a_n\end{align*}$ .
      ::Reindex {n=13nn+4}是表_n=7an的合计。
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  12. Reindex $\begin{align*}\sum\limits_{n=1}^\infty \frac{2^{n-1}}{n!}\end{align*}$  to be a sum of the form $\begin{align*}\sum\limits_{n=0}^\infty a_n\end{align*}$ .
      ::Reindex @n=12n-1n! 以作为表格_n=0an的总和 。
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  13. Write a series summing:
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      1. the even summands of $\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{n}\end{align*}$ , i.e. $\begin{align*}\sum\limits_{n=1}^\infty a_n=\frac{1}{2}+\frac{1}{4}+\dots\end{align*}$ .
         ::+N=11n的偶数总和, 即: @n=1an=12+14+...
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      2. the odd summands of $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{n}\end{align*}$ , i.e. $\begin{align*}\sum\limits_{n=1}^\infty a_n=\frac{1}{3}+\frac{1}{5}+\dots\end{align*}$ .
         ::奇数总和为n=11n, 即n=1an=13+15+...
      
      ::写一系列曲目: n=11n, 即n=1an=12+14+.... n=11n, 即n=1an=13+15+.
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  14. Reindex the series $\begin{align*}\sum\limits_{n=1}^\infty\left[\frac{2}{n^2}+7\left(\frac{1}{3}\right)^{n+3}\right]\end{align*}$ to start at an index of 10.
      ::将序列n=1[2n2+7(13n+3)重新索引,从10的指数开始。
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  15. Reindex the series $\begin{align*}\sum\limits_{n=5}^\infty \frac{n^2+3n+2}{4n^2+1}\end{align*}$  to start at an index of 1.
      ::重新索引 = 5n2+3n+24n2+1 序列,从 1 的指数开始。

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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 。