13.8 部分总和和和确定无限几何序列的总和

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• 选择节无限几何序列部分总和和和计算总和

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  无限几何序列部分总和和和计算总和

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  Suppose you receive an annuity, an investment with an initial lump-sum premium that pays out fixed amounts, in perpetuity (that is, forever). The payment is $100 once a year at the end of the year. Receiving $100 a year from now is worth less than receiving $100 now, because you cannot invest the money until you receive it. The present value of $100 one year in the future is $\begin{array}{r}\frac{100}{1+r},\end{array}$  where $\begin{array}{r}r\end{array}$  is the annual interest rate. Similarly, the present value of a payment of $100 two years in the future is $\begin{array}{r}\frac{100}{(1+r{)}^2},\end{array}$  and so on . Therefore, the present value of receiving $100 per year forever is  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{100}{(1+r{)}^n}\end{array}$  . ¹
  ::假设您收到年金,即初始一次总付的一次性保险费投资,以永久支付固定数额(即永久),年终时每年支付100美元。从现在起每年领取100美元,价值低于100美元,因为直到你收到这笔钱之前,你现在还不能投资这笔钱。未来每年100美元的现值是1001+r,年利率是r。同样,未来两年支付100美元的现值为100(1+1+r)2,等等。因此,每年领取100美元的现值是100美元=100+1美元(1+r)n。

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  This is an example of an infinite series. We discuss series of this type below. 
  ::这是无穷无尽的系列的例子。下面我们讨论这种类型的系列。

  

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  Partial Sums
  ::部分总和

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  An infinite series is a series with an infinite number of terms. In other words, the value of $\begin{array}{r}n\end{array}$ increases without bound as shown in the series below.
  ::无限系列是一系列,用无限数的术语。换句话说,n的增加值没有约束,如下文系列所示。

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  $$\begin{array}{rl}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(3n+1)& =4+7+10+13+\dots \\ \underset{n=1}{\overset{\mathrm{\infty }}{\sum }}4(2{)}^{n-1}& =4+8+16+32+\dots \\ \underset{n=1}{\overset{\mathrm{\infty }}{\sum }}8{\left(\frac{1}{2}\right)}^{n-1}& =8+4+2+1+\frac{1}{2}+\dots \end{array}$$

  
  ::n=1( 3n+1) = 4+7+10+13+... n= 14(2)n- 1= 4+8+16+32+... n=18( 12n- 1=8+4+2+1+12+...

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  These sums continue forever and can increase without bound. As we add more and more terms if the sum gets close to— that is, "approaches"— a particular number, it is said to converge . If it grows without bound, it is said to diverge . 
  ::这些数量将永远持续下去,并且可以不受约束地增加。 当我们增加越来越多的条件时,如果这个数量接近一个特定的数量,也就是“近似 ” , 据说它会趋同。 如果它没有约束地增长,它就会变异。

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  Since we cannot always find the values of these series by adding all the terms, we can analyze their behavior by observing the patterns of  their .
  ::由于我们无法总是通过添加所有术语来找到这些系列的价值观,我们可以通过观察它们的模式来分析它们的行为。

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     Partial Sums of a
  ::部分总和 a

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  For an infinite sequence of terms, $\begin{array}{r}{a}_i\end{array}$ , the partial sums are
  ::对于无限顺序的术语,Ai,部分金额是

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  $$\begin{array}{rl}{S}_1& =a_1\\ S_2& =a_1+a_2\\ S_3& =a_1+a_2+a_3\\ & ⋮\\ S_n& =a_1+a_2+a_3+\dots +a_n\end{array}$$

  
  $\begin{array}{r}{S}_n\end{array}$  is the nth partial sum. 
  ::S1=a1S2=a1+a2S3=a1+a2+a3=Sn=a1+a2+a3+...+anSn为nth部分总和。

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  The sequence $\begin{array}{r}{S}_1,S_2,S_3,\dots ,S_n,\dots \end{array}$ is the .  
  ::S1,S2,S3,S3的顺序是

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  We can look at a sequence of partial sums to observe the behavior of an  infinite sum. Let's consider how to find partial sums below. 
  ::我们可以看一连串部分金额 来观察无限金额的行为。让我们考虑如何在下面找到部分金额。

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  Example 1
  ::例1

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  Find the 1st five partial sums of $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(2n-1)\end{array}$ and make an observation about the sum of the infinite series.
  ::查找第 1 个部分和%n=1 ( 2n-1) , 并观察无限序列的总和 。

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  Solution: The 1st five partial sums are $\begin{array}{r}{S}_1,S_2,S_3,S_4\end{array}$ and $\begin{array}{r}{S}_5\end{array}$ . To find each of these sums, we need to find the 1st five terms of the sequence. The 1st five terms of the sequence are the 1st five odd numbers, 1, 3, 5, 7, 9. Now we can find the partial sums:
  ::解答:前五个部分金额是S1、S2、S3、S4和S5。为了找到其中每一个金额,我们需要找到顺序的第五个条件。顺序的第五个条件是第五个奇数,1、3、5、7、9。 现在我们可以找到部分金额:

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  $$\begin{array}{rl}{S}_1& =a_1=1\\ S_2& =a_1+a_2=1+3=4\\ S_3& =a_1+a_2+a_3=1+3+5=9\\ S_4& =a_1+a_2+a_3+a_4=1+3+5+7=16\\ S_5& =a_1+a_2+a_3+a_4+a_5=1+3+5+7+9=25\end{array}$$

  
  ::S1=a1=1S2=1S2=a1+a2=1+a2+3=1+3=3=4S3=a1+a2+3+3+5=9S4=a1+a2+a3+a4=1+3+3+5+7=16S5=a1+a2+a3+a3+a4+a5=1+3+5+5+9=25

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  Notice that each sum can also be found by adding the $\begin{array}{r}{n}^{th}\end{array}$ term to the previous sum,  $\begin{array}{r}{S}_n=S_{n-1}+a_n\end{array}$ . For example,  $\begin{array}{r}{S}_5=S_4+a_5=16+9=25\end{array}$ .
  ::通知中说,在上一个总和Sn=Sn=Sn-1+an中加上n=Sn=S4+a5=16+9=25,即可找到每一笔总和。

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  The sequence of the 1st five partial sums is 1, 4, 9, 16, 25. This pattern will continue and the terms will continue to grow without bound. In other words, the partial sums continue to grow and the infinite sum cannot be determined as it is infinitely large. The sum diverges .
  ::第1五个部分总和的顺序是1、4、9、16、25。这种模式将继续下去,术语将无约束地继续增长。换句话说,部分总和将继续增长,无法确定无限总和,因为它是无限大的。总和是不同的。

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  Example 2
  ::例2

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  Find the 1st five partial sums of $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}{\left(\frac{1}{2}\right)}^{n-1}\end{array}$ and make an observation about the infinite series.
  ::查找第 1 个部分和%n=1( 12)n- 1, 并对无限序列进行观察 。

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  Solution: The 1st five terms of this sequence are powers of $\begin{array}{r}\frac{1}{2}\end{array}$ : $\begin{array}{r}1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16}.\end{array}$  The partial sums are
  ::解决办法:本序列中第五个条件的第一条是12:1,12,14,18,116的权力。

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  $$\begin{array}{rl}{S}_1& =1\\ S_2& =1+\frac{1}{2}=1\frac{1}{2}\\ S_3& =1+\frac{1}{2}+\frac{1}{4}=1\frac{3}{4}\\ S_4& =1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1\frac{7}{8}\\ S_5& =1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\frac{15}{16}.\end{array}$$

  
  ::S1=1S2=1+12=112S3=1+12+14=134S4=1+12+14+18=178S5=1+12+14+18+116=11516。

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  Consider what happens with each subsequent term: We start with 1 and add $\begin{array}{r}\frac{1}{2},\end{array}$ putting us halfway between 1 and 2. Then we add $\begin{array}{r}\frac{1}{4}\end{array}$ , putting us halfway between 1.5 and 2. Each time we add another term, we are cutting the distance between our current sum and 2 in half. If this pattern is continued, we will get ever closer to 2 but never actually reach 2. Therefore, the sum of this sequence or series is said to converge to or "approach" 2.
  ::想想接下来的每个任期会怎样:我们从1开始,再加12,让我们在1到2之间走一半,再加14,让我们在1.5到2之间走一半,每次我们再加一个任期,我们就会缩短我们目前和2之间的距离。如果这种模式继续下去,我们就会越来越接近2,但实际上永远不会达到2。 因此,这个序列或系列的总和据说会与或“接近2”汇合或“接近2”。

   

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  Example 3
  ::例3

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  Find the 1st five partial sums, the 50th, the 100th, and the 500th partial sums of $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{n}\end{array}$ , the "harmonic series," and make an observation about the infinite series.
  ::寻找第1五个部分总和, 第50个, 第100个,和第500个部分总和 n=11n, "和谐序列", 并对无限序列进行观察。

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  Solution:  Since we are going to sum many terms, let's use a calculator to find the sums below. (These are the commands with a TI-83/84. You can also find these sums using Desmos.)
  ::解决方案:既然我们要计算许多术语,让我们用计算器来找到以下的金额。 (这些是使用 TI- 83/84 的命令。 您也可以使用 Desmos 找到这些金额 。)

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  $$\begin{array}{rl}{S}_1& =sum(seq(1/x,x,1,1))=1\\ S_2& =sum(seq(1/x,x,1,2))=1.5\\ S_3& =sum(seq(1/x,x,1,3))=1.833\\ S_4& =sum(seq(1/x,x,1,4))=2.083\\ S_5& =sum(seq(1/x,x,1,5))=2.283\end{array}$$

  
  ::S1=Ssum(1,x,x,1,1)=1,S2=sum(seq(1/x,x,1,2))=1.5S3=sum(seq(1/x,x,x,1,3))=1.833S4=sum(seq(1/x,x,x,1,4))=2.083S5=sum(seq(1,x,x,1,5)=2.283

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  In this series, the behavior is not quite as clear. Consider some additional partial sums:
  ::在这一系列行为中,行为并不十分明确。

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  $$\begin{array}{rl}{S}_{50}& =4.499\\ S_{100}& =5.187\\ S_{500}& =6.793\end{array}$$

  
  ::S50=4.4999S100=5.187S500=6.793

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  In this case, the partial sums do not seem to have a bound. They will continue to grow. This series diverges. 
  ::在这种情况下,部分金额似乎没有约束性,它们将继续增长。

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  by Mathispower4u demonstrates how to graph the sequence of partial sums on a TI-83/84 to see if the sequence of partial sums converges or diverges. 
  ::通过 Mathispower4u 演示如何用TI-83/84 绘制部分总和的顺序图,以确定部分总和的顺序是否趋同或不同。

   

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  Example 4
  ::例4

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  Find the 1st five partial sums of the infinite series below, and additional partial sums if needed, to determine if  the infinite series converges or diverges.
  ::查找以下无限序列的第一和第二部分总和,必要时查找额外的部分总和,以确定无限序列是否趋同或不同。

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  a. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}4{\left(\frac{3}{2}\right)}^{n-1}\end{array}$
  ::a. a. n=1 4(32)n-1

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  b. $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}500{\left(\frac{2}{3}\right)}^{n-1}\end{array}$
  ::b. b. n=1 =1 500(23)n-1

   

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  Solution:
  ::解决方案 :

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  a. $\begin{array}{r}{S}_1=4;S_2=10;S_3=19;S_4=32.5;S_5=\mathrm{52.75.}\end{array}$ The partial sums are growing with increasing speed, and thus the infinite series will have no bound. This series diverges.
  ::a. S1=4;S2=10;S3=19;S4=32.5;S5=52.75。部分总和正在以不断加快的速度增长,因此无限序列将没有约束。这一序列各不相同。

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  b. $\begin{array}{r}{S}_1=500;S_2=833.333;S_3=1,055.556;S_4=1,203.704;S_5=1,\mathrm{302.469.}\end{array}$ Here the sums seem to be growing by smaller amounts each time. Look at additional partial sums to see if there is an apparent upper limit to their growth. $\begin{array}{r}{S}_{50}=1,499.9999\dots \approx 1,500;S_{100}\approx 1,500.\end{array}$ The sum is approaching 1,500 and thus the infinite series has a finite sum. This series converges.
  ::b. S1=500;S2=833.333;S3=1,055.556;S4=1,203.704;S5=1,302.469。总金额似乎每次以较小数量增长。看看额外的部分金额,看其增长是否有明显的上限。S50=1,499.99...1,500;S1001,500。总和接近1,500,因此无限序列有一定数量。这个序列相交。

   

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  Sum of an Infinite Geometric Sequence
  ::无限几何序列总和

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  Now we will focus on geometric series, because we can find a formula for the sum of an infinite geometric sequence. Let's look at the partial sums of the infinite geometric series below:
  ::现在我们将集中关注几何序列, 因为我们能找到一个公式, 计算一个无限几何序列的总和。 让我们来看看以下无限几何序列的局部总和 :

  Series	$\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}3(1{)}^{n-1}\end{array}$	$\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}10{\left(\frac{3}{4}\right)}^{n-1}\end{array}$	$\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}5{\left(\frac{6}{5}\right)}^{n-1}\end{array}$	$\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}(\text{-}2{)}^{n-1}\end{array}$	$\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2{\left(\text{-}\frac{1}{3}\right)}^{n-1}\end{array}$
  $\begin{array}{r}{S}_5\end{array}$	15	30.508	37.208	11	1.506
  $\begin{array}{r}{S}_{10}\end{array}$	30	37.747	129.793	$\begin{array}{r}\text{-}341\end{array}$	$\begin{array}{r}\approx \end{array}$  1.5
  $\begin{array}{r}{S}_{50}\end{array}$	150	$\begin{array}{r}\approx \end{array}$  40	227,485.954	$\begin{array}{r}\text{-}3.753\times {10}^{14}\end{array}$	$\begin{array}{r}\approx \end{array}$  1.5
  $\begin{array}{r}{S}_{100}\end{array}$	300	$\begin{array}{r}\approx \end{array}$  40	2,070,449,338	$\begin{array}{r}\text{-}4.226\times {10}^{29}\end{array}$	$\begin{array}{r}\approx \end{array}$  1.5

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  From the table above, we can see that the two infinite geometric series that have a finite sum are $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}10{\left(\frac{3}{4}\right)}^{n-1}\end{array}$ and $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2{\left(-\frac{1}{3}\right)}^{n-1}\end{array}$ . The two sequences both have a common ratio, $\begin{array}{r}r\end{array}$ , such that $\begin{array}{r}\left|r\right|<1\end{array}$ or $\begin{array}{r}\text{-}1<r<1\end{array}$ . This will be a necessary condition for us to find a sum of an infinite geometric sequence. 
  ::从上表可以看出,两个无限的几何序列的数值有限,它们是n=110(34)n-1和n=12(-13)n-1。 这两个序列都有共同比率,r,例如r1或-1<r>1。 这将是我们找到无限几何序列之和的必要条件。

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  Now consider the formula for the sum of a finite geometric sequence,  $\begin{array}{r}{S}_n=\frac{a_1\left(1-r^n\right)}{1-r}\end{array}$ . What happens to $\begin{array}{r}{r}^n\end{array}$ if we let $\begin{array}{r}n\end{array}$ get very large for an $\begin{array}{r}r,\end{array}$ such that $\begin{array}{r}\left|r\right|<1\end{array}$ ? Let's take a look at some examples.
  ::现在考虑一个限定几何序列总和的公式, Sn=a1(1-rn)1-r。 如果我们让一个 r 变得非常大, 那么这样, r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\的\\\\\\\\\\\Sn- \ \ Sn\\\\ \\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

  $\begin{array}{r}r\end{array}$ values	$\begin{array}{r}{r}^5\end{array}$	$\begin{array}{r}{r}^{25}\end{array}$	$\begin{array}{r}{r}^{50}\end{array}$
  $\begin{array}{r}\frac{5}{6}\end{array}$	$\begin{array}{r}0.40188\end{array}$	$\begin{array}{r}0.01048\end{array}$	$\begin{array}{r}0.00011\end{array}$
  $\begin{array}{r}\text{-}\frac{4}{5}\end{array}$	$\begin{array}{r}\text{-}0.32768\end{array}$	$\begin{array}{r}\text{-}0.00378\end{array}$	$\begin{array}{r}0.00001\end{array}$
  $\begin{array}{r}\text{-}\frac{1}{3}\end{array}$	$\begin{array}{r}\text{-}0.00412\end{array}$	$\begin{array}{r}\text{-}1.18024\times {10}^{\text{-}12}\end{array}$	$\begin{array}{r}1.39296\times {10}^{\text{-}24}\end{array}$

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  This table shows that when $\begin{array}{r}\left|r\right|<1\end{array}$ , $\begin{array}{r}{r}^n\approx 0\end{array}$  for large values of $\begin{array}{r}n\end{array}$ . Therefore, for the infinite geometric series in which $\begin{array}{r}\left|r\right|<1,S_{\mathrm{\infty }}=\frac{a_1\left(1-r^{\mathrm{\infty }}\right)}{1-r}\approx \frac{a_1\left(1-0\right)}{1-r}=\frac{a_1}{1-r}.\end{array}$
  ::本表显示,当 n. 的大值为 r1, r0时, 则表示无限几何序列中的 r1,Sa1(1-r1)1-ra1(1-0)1-r=a11-r。

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     Sum of an Infinite Geometric Sequence
  ::无限几何序列总和

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  Given  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}{a}_1{r}^{n-1}\end{array}$  with  $\begin{array}{r}|r|<1\end{array}$ , the sum $\begin{array}{r}{S}_{\mathrm{\infty }}\end{array}$  of all of the terms in the geometric sequence is
  ::鉴于“n”=1a1rn-1加上“r”1, 几何序列中所有术语的和 S=

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  $$\begin{array}{r}{S}_{\mathrm{\infty }}=\frac{a_1}{1-r}.\end{array}$$

  
  ::萨11 -r.

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  Note we cannot use this formula if  $\begin{array}{r}|r|\ge 1\end{array}$ . In this case, the sum diverges.  
  ::请注意, 如果“r1” , 我们就不能使用此公式。 在这种情况下, 求和是不同的 。

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  Example 5
  ::例5

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  Evaluate  the geometric series if possible:  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}100{\left(\frac{8}{9}\right)}^{n-1}\end{array}$ .
  ::如果可能,评估几何序列:n=1100(89)n-1。

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  Solution: Using the formula with $\begin{array}{r}{a}_1=100\end{array}$ , $\begin{array}{r}r=\frac{8}{9}\end{array}$ , we get $\begin{array}{r}{S}_{\mathrm{\infty }}=\frac{100}{1-\frac{8}{9}}=\frac{100}{\frac{1}{9}}=900.\end{array}$
  ::解答:使用a1=100,r=89的公式,我们得到S1001-89=10019=900。

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  Example 6
  ::例6

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  Evaluate the geometric series if possible:  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}9{\left(\frac{4}{3}\right)}^{n-1}\end{array}$ .
  ::如果可能,评估几何序列:n=19(43)n-1。

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  Solution: In this case, $\begin{array}{r}\left|r\right|=\frac{4}{3}>1\end{array}$ , therefore the sum is infinite and cannot be determined.
  ::解答:在此情况下,`r43>1,因此,总和是无限的,无法确定。

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  Example 7
  ::例7

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  Evaluate the geometric series if possible:  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}5(0.99{)}^{n-1}\end{array}$ .
  ::如有可能,评估几何序列:n=15(0.99n-1)。

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  Solution: In this case,  $\begin{array}{r}{a}_1=5\end{array}$ and $\begin{array}{r}r=0.99\end{array}$ , so $\begin{array}{r}{S}_{\mathrm{\infty }}=\frac{5}{1-0.99}=\frac{5}{0.01}=500\end{array}$ .
  ::解决办法:在这种情况下,a1=5和r=0.99,S51-0.99=50.01=500。

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  Example 8
  ::例8

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  Evaluate the geometric series if possible:  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}3(-1{)}^{n-1}\end{array}$ .
  ::如果可能的话,评估几何序列:n=13(-1)-n-1。

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  Solution:   $\begin{array}{r}\left|r\right|=\left|-1\right|=1\ge 1\end{array}$ , therefore the infinite sum does not converge. If we observe the behavior of the 1st few partial sums, we can see that they oscillate between 0 and 3.
  ::解答: @r111111, 因此无限之和无法趋同。 如果我们观察一小部分之和的行为, 我们可以看到它们在0到3之间徘徊。

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  $$\begin{array}{rl}{S}_1& =3\\ S_2& =0\\ S_3& =3\\ S_4& =0\end{array}$$

  
  ::S1=3S2=0S3=3S4=0

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  This pattern will continue so there is no determinable sum for the infinite series.
  ::这种模式将继续下去,所以无限序列没有可确定的总和。

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  by Mathispower4u demonstrates how to use the formula for the sum of an infinite geometric sequence. 
  ::由 Mathispower4u 演示如何使用公式来计算无限几何序列的总和。

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  Example 9 
  ::例9

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  Evaluate the geometric series  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{100}{(1+r{)}^n}\end{array}$  where  $\begin{array}{r}r\end{array}$  is 5%.
  ::评估几何序列 n=1100(1+r)n,r为5%。

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  Solution: First, we need to rewrite the sum so that it fits with the formula. 
  ::解决方案:首先,我们需要重写这笔金额,使之与公式相符。

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  $$\begin{array}{rl}& =\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{100}{(1+0.05{)}^{n-1}}\cdot \frac{1}{1+0.05}\\ & =\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{100}{1+0.05}\cdot (\frac{1}{1+0.05}{)}^{n-1}\end{array}$$

  Since $\begin{array}{r}|\frac{1}{1.05}|<1\end{array}$ , we can use the sum formula.
  ::n=1100(1+0.05)n-111+0.05n=11001+0.05(11+0.05)n-1Since11.051,我们可以使用总和公式。

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  $$\begin{array}{rl}{S}_{\mathrm{\infty }}& =\frac{\frac{100}{1+0.05}}{1-\frac{1}{1.05}}\\ & =\frac{100}{1.05}\cdot \frac{1}{0.0476}\\ & =\frac{100}{0.05}\\ & =2,000\end{array}$$

  The present value is $2,000.   
  ::S1001+0.051-11.05=1001.0510.0476=1000.05=2 000 现值为2 000美元。

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  Feature: Hot Topics
  ::特专题:热题

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  by Deirdre Mundy
  ::由Deirdre Mundy 编辑

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  Since the 1800s, scientists have used infinite series to explain how heat moves between objects. In the past, the science of heat transfer mostly involved understanding how common materials worked. Now scientists use the mathematics of heat to develop new materials. Using the mathematics of heat transfer, scientists may find a way to combat global warming.
  ::自1800年代以来,科学家们用无限的系列来解释不同物体之间的热量移动方式。 过去,热传输科学主要涉及了解常见材料是如何起作用的。 如今,科学家们利用热量数学来开发新材料。 科学家们利用热量转移的数学,可以找到防止全球变暖的方法。

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  Solar Problems, Heated Solutions
  ::太阳能问题,加热解决方案

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  Many environmentalists cite solar power as a way to combat global warming. Photovoltaic solar panels turn energy from the sun into electricity without adding carbon dioxide to the atmosphere. It's true that PV solar panels are a clean, green energy source, but they have a major flaw: Once the sun stops shining, the panels stop creating electricity. That means that if consumers want electricity 24 hours a day, solar cells aren't the answer.
  ::许多环保主义者将太阳能作为对抗全球变暖的一种方法。光伏太阳能电池板将太阳能从太阳变成电,而不给大气增加二氧化碳。光伏太阳能电池板确实是一个清洁绿色能源,但它们有一个重大缺陷:太阳一停光,电池板就停止发电。这意味着如果消费者每天24小时都想要电力,太阳能电池板就不是答案。

  

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  There's a different kind of solar plant that can create electricity 24 hours a day. A concentrated solar power (CSP) plant uses mirrors to redirect sunlight towards long tubes. The tubes are full of a liquid called a heat transfer fluid. These liquids can absorb heat from the sun and then release it later, to heat water. The boiling water becomes steam, which turns a turbine. The turbine generates electricity.
  ::一种不同的太阳能电厂,可以每天24小时发电。一个集中的太阳能电厂用镜子将阳光引向长管。管里装满了一种液体,叫做热传导液。这些液体可以吸收太阳的热量,然后释放出来,用热水。沸水会变成蒸汽,转动涡轮机。涡轮机能发电。

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  Scientists have engineered heat transfer fluids that can hold a lot of heat for a long time. They've also created insulated storage tanks for the fluids. That means that CSP plants can store hot liquid overnight, and use it to make steam all night long.
  ::科学家们设计了热传导液, 能够长期保持大量热量。 他们也为液体制造了绝缘储油罐。 这意味着CSP工厂可以隔夜储存热液, 并用它做整夜蒸汽。

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  As scientists develop better heat transfer fluids, they'll be able to make CSP plants even more efficient. This means they'll be able to use fewer mirrors and less space to generate the same amount of electricity, diminishing the environmental impact of the plants.
  ::随着科学家们开发出更好的热传导流体, 他们将能够提高CSP工厂的效率。 这意味着他们可以使用更少的镜像和更少的空间 来产生同样数量的电力, 减少这些工厂对环境的影响。

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  Dow Chemical Company produces a silicone-based heat transfer fluid that can work at temperatures around 800 degrees Fahrenheit. Meanwhile, researchers have also created a nanoparticle that can turn 90% of the light energy that strikes it into heat energy. With advances like these, it may soon be possible to power the world with sunlight.
  ::道氏化学公司生产的硅酮热传导液可以在华氏800度左右的温度下工作。 与此同时,研究人员还制造了纳米粒子,可以将90%的光能转化为热能。 有了这样的进步,也许不久就能用阳光为世界提供动力。

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  by the U.S. Department of Energy discusses concentrating solar power. 
  ::美国能源部讨论太阳能发电。

   

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  Summary
  ::摘要

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  • An infinite series is a series with an infinite number of terms.
    ::无限的系列是一个有无限数术语的系列。
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  • If the sum gets close to—that is, "approaches"— a particular number, it is said to converge. If it grows without bound, it is said to diverge.
    ::如果总和接近——即“近似”——一个特定的数字,据说它会趋同,如果它不受约束地生长,它就会变异。
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  • Since we cannot always find these series by adding all the terms, we can analyze their behavior by observing the patterns of their partial sums. Given $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}{a}_1{r}^{n-1}\end{array}$  with $\begin{array}{r}|r|<1\end{array}$ , the sum $\begin{array}{r}S\infty \end{array}$  of all the terms in the geometric sequence is  $\begin{array}{r}{S}_{\mathrm{\infty }}=\frac{a_1}{1-r}.\end{array}$  
    ::由于我们无法总是通过添加所有术语来找到这些系列, 我们可以通过观察其部分总和的规律来分析它们的行为模式。 鉴于 n=1 a1 1 和 1, 几何序列中所有术语的总和 S1 是 Sa11-r。

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  Review
  ::回顾

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  Find the 1st five partial sums and additional partial sums as needed to discuss the behavior of each infinite series. 
  ::找出第1五个部分总和 和额外的部分总和 以讨论每一无限系列的行为

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  1.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}5{\left(\frac{1}{2}\right)}^{n-1}\end{array}$
  ::1. =1=5(12)-1

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  2.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2{\left(\frac{3}{4}\right)}^{n-1}\end{array}$
  ::2. =1=1 (2) (3)-1

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  3.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}10(0.9{)}^{n-1}\end{array}$
  ::3.n=110(0.9.9)n-1

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  4.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}8(1.03{)}^{n-1}\end{array}$
  ::4.n=18(1.03n-1)

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  5.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2}n\end{array}$
  ::5.n=112n

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  6.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{10}{n}\end{array}$
  ::6.n=110n

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  7.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2}{\left(\frac{3}{4}\right)}^{n-1}\end{array}$
  ::7.n=112(34n-1)

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  8.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{n^2}\end{array}$
  ::8.n=11n2

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  9.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}6(0.1{)}^{n-1}\end{array}$
  ::9.n=16(0.1)n-1

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  10.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}0.01n+5\end{array}$
  ::10.=1 0.01n+5

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  11.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2{\left(\frac{7}{8}\right)}^{n-1}\end{array}$
  ::11.n=1=1 (2(78)-1

   

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  Evaluate the infinite geometric series, if possible.
  ::如果可能的话,评估无限几何序列。

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  12.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}5{\left(\frac{2}{3}\right)}^{n-1}\end{array}$
  ::12. =1=5(23)-1

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  13.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{10}{\left(\text{-}\frac{4}{3}\right)}^{n-1}\end{array}$
  ::13.n=1 110(- 43n-1)

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  14.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2{\left(\text{-}\frac{1}{3}\right)}^{n-1}\end{array}$
  ::14. =1=1(2-13-13-1)

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  15.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}8(1.1{)}^{n-1}\end{array}$
  ::15.n=18(1.1)n-1

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  16.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}6(0.4{)}^{n-1}\end{array}$
  ::16. =1 6(0.4)n-1

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  17.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2}{\left(\frac{3}{7}\right)}^{n-1}\end{array}$
  ::17.n=112(37n-1)

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  18.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{5}{3}{\left(\frac{1}{6}\right)}^{n-1}\end{array}$
  ::18.n=153(16.n-1)

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  19.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{5}(1.05{)}^{n-1}\end{array}$
  ::19.n=115(1.05n-1)

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  20.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{4}{7}{\left(\frac{6}{7}\right)}^{n-1}\end{array}$
  ::20.n=147(67n-1)

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  21.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}15{\left(\frac{11}{12}\right)}^{n-1}\end{array}$
  ::21.n=115(1112)n-1

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  22.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}0.01{\left(\frac{3}{2}\right)}^{n-1}\end{array}$
  ::22.n=10.01(32°n-1)

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  23.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}100{\left(\frac{1}{5}\right)}^{n-1}\end{array}$
  ::23.n=1100(15毫-1)

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  24.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{2}{\left(\frac{5}{4}\right)}^{n-1}\end{array}$
  ::24.n=112(54n-1)

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  25.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}2.5(0.85{)}^{n-1}\end{array}$
  ::25.n=12.5(0.85)n-1

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  26.  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}-3{\left(\frac{9}{16}\right)}^{n-1}\end{array}$
  ::26.n=13 (916)n-1

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  Explore More
  ::探索更多

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  1.a. Which of the series above are arithmetic? Do any of them have a finite sum? Can you explain why?
  ::1.a. 上面哪个系列是算术?它们中是否有一个有一定数目的?你能解释一下为什么吗?

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  b. Which of the series above are geometric? Do any of them have a finite sum? Can you explain why?
  ::b. 以上哪一组是几何数组?其中是否有一个有一定的数值?你能解释一下为什么吗?

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  c. What is the difference between the series that converge and diverge?
  ::c. 趋同和相异的系列之间有什么区别?

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  d. Make a conjecture about the converging series in this problem set.
  ::d. 对这一组问题中汇集的系列进行推测。

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  2. In the 18th century, mathematician Leonhard Euler solved one of the foremost infinite series problems of his day by examining the series $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{2}{n(n+1)}\end{array}$ . ²   Find the 1st five partial sums of this series and make an observation about the infinite series.
  ::2. 在18世纪,数学家Leonhard Euler通过审查系列丛书"n=12n(n+1),2 解决了他这一天最大的无限系列问题之一。

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  3. Euler also solved the Basel problem, ² which is to find the series  $\begin{array}{r}\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}\frac{1}{n^2}\end{array}$ .    Find the 1st five partial sums of this series and make an observation about the infinite series.
  ::3. Euler还解决了巴塞尔问题,2 即寻找序列 n=11n2. 找到该系列的第一和第二部分总和,并对无限序列进行观察。

   

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  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

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  Please see the Appendix.
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try this interactive that reinforces the concepts explored in this section:
  ::尝试这一互动,强化本节所探讨的概念:

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  References
  ::参考参考资料

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  1. "Annuity (American)," last edited March 18, 2017, 
  ::1. “年金(美国)”,2017年3月18日编辑,

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  2. "An Infinite Series of Surprises," by C.J. Sangwin, submitted December 1, 2001, 
  ::2. C.J.Sangwin于2001年12月1日提交,