11.12 寻找无限几何系列的总数

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  查找无限几何序列的总和

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  Your task as Agent Infinite Geometric Series , should you choose to accept it, is to find the sum of the geometric series $\begin{align*}\sum \limits_{n=1}^{\infty}3 \left(\frac{1}{3}\right)^{n-1}\end{align*}$ .
  ::您作为Infinite几何系列特工的任务, 如果您选择接受的话, 是找到几何序列的和数 n=1\%3(13)n- 1 。

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  Sum of Infinite Geometric Series
  ::无限几何系列总和

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  We have explored of various infinite series and observed their behavior as $\begin{align*}n\end{align*}$ became large to see if the sum of the infinite series was finite. Now we will focus our attention on geometric series. Look at the partial sums of the infinite geometric series below:
  ::我们探索了无穷无尽的系列,观察了它们作为n大的行为,看无限系列的总和是否是有限的。现在,我们将集中关注几何系列。看看无限几何系列以下的部分总和:

  Series	$\begin{align*}\sum \limits_{n=1}^{\infty}3(1)^{n-1}\end{align*}$	$\begin{align*}\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}\end{align*}$	$\begin{align*}\sum \limits_{n=1}^{\infty}5 \left(\frac{6}{5}\right)^{n-1}\end{align*}$	$\begin{align*}\sum \limits_{n=1}^{\infty}(-2)^{n-1}\end{align*}$	$\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}$
  $\begin{align*}S_5\end{align*}$	15	30.508	37.208	11	1.506
  $\begin{align*}S_{10}\end{align*}$	30	37.747	129.793	$\begin{align*}-341\end{align*}$	1.5
  $\begin{align*}S_{50}\end{align*}$	150	40	227485.954	$\begin{align*}-3.753 \times 10^{14}\end{align*}$	1.5
  $\begin{align*}S_{100}\end{align*}$	300	40	2070449338	$\begin{align*}-4.226 \times 10^{29}\end{align*}$	1.5

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  From the table above, we can see that the two infinite geometric series which have a finite sum are $\begin{align*}\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}\end{align*}$ and $\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}$ . The two series both have a common ratio, $\begin{align*}r\end{align*}$ , such that $\begin{align*}\left | r \right \vert < 1\end{align*}$ or $\begin{align*}-1 < r < 1\end{align*}$ .
  ::从上表可以看出,有两个无限的几何序列,其数量有限,它们是n=110(34)n-1和n=12(-13)n-1。 这两个序列都有一个共同比率,r,例如1或-1<r<1。

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  Take a look at the formula for the sum of a finite geometric series: $\begin{align*}S_n=\frac{a_1\left(1-r^n\right)}{1-r}\end{align*}$ . What happens to $\begin{align*}r^n\end{align*}$ if we let $\begin{align*}n\end{align*}$ get very large for an $\begin{align*}r\end{align*}$ such that $\begin{align*}\left |r \right \vert <1\end{align*}$ ? Let’s take a look at some examples.
  ::看一下一个限定几何序列总和的公式: Sn=a1(1-rn)1-r。 如果我们让一个非常大的 r 如此大的 r , 那么被撕裂的公式会怎样? 让我们看看一些例子。

  $\begin{align*}r\end{align*}$ values	$\begin{align*}r^5\end{align*}$	$\begin{align*}r^{25}\end{align*}$	$\begin{align*}r^{50}\end{align*}$	$\begin{align*}\ldots\end{align*}$	$\begin{align*}r^n\end{align*}$ or $\begin{align*}r^{\infty}\end{align*}$
  $\begin{align*}\frac{5}{6}\end{align*}$	$\begin{align*}0.40188\end{align*}$	$\begin{align*}0.01048\end{align*}$	$\begin{align*}0.00011\end{align*}$
  $\begin{align*}- \frac{4}{5}\end{align*}$	$\begin{align*}-0.32768\end{align*}$	$\begin{align*}-0.00378\end{align*}$	$\begin{align*}0.00001\end{align*}$
  $\begin{align*}1.1\end{align*}$	$\begin{align*}1.61051\end{align*}$	$\begin{align*}10.83471\end{align*}$	$\begin{align*}117.39085\end{align*}$		keeps growing
  $\begin{align*}- \frac{1}{3}\end{align*}$	$\begin{align*}-0.00412\end{align*}$	$\begin{align*}-1.18024 \times 10^{-12}\end{align*}$	$\begin{align*}1.39296 \times 10^{-24}\end{align*}$

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  This table shows that when $\begin{align*}\left | r \right \vert <1\end{align*}$ , $\begin{align*}r^n=0\end{align*}$ , for large values of $\begin{align*}n\end{align*}$ . Therefore, for the sum of an infinite geometric series in which $\begin{align*}\left | r \right \vert <1, S_{\infty}=\frac{a_1 \left(1-r^n\right)}{1-r}=\frac{a_1\left(1-0\right)}{1-r}=\frac{a_1}{1-r}\end{align*}$ .
  ::本表显示,当 n. 的大值为 r1, r=0时。 因此, 对于无限几何序列的总和, 其中r1,Sa1(1-rn)1-r=a1(1-0)1-r=a11-r。

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  Let's find the sum of the following geometric series, if possible.
  ::如果可能的话,让我们找出以下几何序列的总和。

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  1. $\begin{align*}\sum \limits_{n=1}^{\infty}100 \left(\frac{8}{9}\right)^{n-1}\end{align*}$ .
     ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!

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  Using the formula with $\begin{align*}a_1=100\end{align*}$ , $\begin{align*}r=\frac{8}{9}\end{align*}$ , we get $\begin{align*}S_{\infty}=\frac{100}{1- \frac{8}{9}}=\frac{100}{\frac{1}{9}}=900\end{align*}$ .
  ::使用a1=100, r=89的公式, 我们得到S1001-89=10019=900。

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  2. $\begin{align*}\sum \limits_{n=1}^{\infty}9 \left(\frac{4}{3}\right)^{n-1}\end{align*}$
     ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  In this case, $\begin{align*}\left | r \right \vert=\frac{4}{3}>1\end{align*}$ , therefore the sum is infinite and cannot be determined.
  ::因此,在这种情况下,该总和是无限的,无法确定。

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  3. $\begin{align*}\sum \limits_{n=1}^{\infty}5(0.99)^{n-1}\end{align*}$
     ::=15(0.99-1)

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  In this case $\begin{align*}a_1=5\end{align*}$ and $\begin{align*}r=0.99\end{align*}$ , so $\begin{align*}S_{\infty}=\frac{5}{1-0.99}=\frac{5}{0.01}=500\end{align*}$ .
  ::在这种情况下,a1=5和r=0.99,因此S51-0.99=50.01=500。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find the sum of the geometric series  $\begin{align*}\sum \limits_{n=1}^{\infty}3 \left(\frac{1}{3}\right)^{n-1}\end{align*}$ .
  ::早些时候,有人要求你找到几何序列的总数。 ”n=13(13)n-1。

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  In this case $\begin{align*}a_1=3\end{align*}$ and $\begin{align*}r=\frac{1}{3}\end{align*}$ , so $\begin{align*}S_{\infty}=\frac{3}{1-\frac{1}{3}}=\frac{3}{\frac{2}{3}}=\frac{9}{2}=4.5\end{align*}$ .
  ::在这种情况下,a1=3和r=13,所以S31-13=323=92=4.5。

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  Find the sums of the following infinite geometric series, if possible.
  ::如果可能的话,找到以下无限几何序列的数值。

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  Example 2
  ::例2

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  $\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{9} \left(- \frac{3}{2}\right)^{n-1}\end{align*}$
  ::=119(-32-32)-1

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  $\begin{align*}\left | r \right \vert=\left |- \frac{3}{2} \right \vert=\frac{3}{2}>1\end{align*}$ so the infinite sum does not exist .
  ::r3232>1 所以无限之和不存在

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  Example 3
  ::例3

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  $\begin{align*}\sum \limits_{n=1}^{\infty}4 \left(\frac{7}{8}\right)^{n-1}\end{align*}$
  ::=14(78-1)

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  $\begin{align*}a_1=4\end{align*}$ and $\begin{align*}r=\frac{7}{8}\end{align*}$ so $\begin{align*}S_{\infty}=\frac{4}{1- \frac{7}{8}}=\frac{4}{\frac{1}{8}}=32\end{align*}$ .
  ::a1=4,r=78,S41-78=418=32。

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  Example 4
  ::例4

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  $\begin{align*}\sum \limits_{n=1}^{\infty}3(-1)^{n-1}\end{align*}$
  ::=13(-1)-1

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  $\begin{align*}\left | r \right \vert=\left |-1 \right \vert=1 \ge 1\end{align*}$ , therefore the infinite sum does not converge . If we observe the behavior of the first few partial sums we can see that they oscillate between 0 and 3.
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  $$\begin{align*}S_1&=3 \\ S_2&=0 \\ S_3&=3 \\ S_4&=0\end{align*}$$
  ::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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  Review
  ::回顾

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  Find the sums of the infinite geometric series, if possible.
  ::如果可能的话,找到无限几何序列的数值。

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  1. $\begin{align*}\sum \limits_{n=1}^{\infty} 5 \left(\frac{2}{3}\right)^{n-1}\end{align*}$
     ::=1=5(23)-1
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  2. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{10} \left(- \frac{4}{3}\right)^{n-1}\end{align*}$
     ::=1 110 (- 43n-1)
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  3. $\begin{align*}\sum \limits_{n=1}^{\infty} 2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}$
     ::=1=1 (2- 13- 13- 1)
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  4. $\begin{align*}\sum \limits_{n=1}^{\infty} 8(1.1)^{n-1}\end{align*}$
     ::=1 =8(1.1)n-1
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  5. $\begin{align*}\sum \limits_{n=1}^{\infty} 6(0.4)^{n-1}\end{align*}$
     ::=1 =6(0.4)n-1
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  6. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{3}{7}\right)^{n-1}\end{align*}$
     ::=112(37°-1)
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  7. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{5}{3} \left(\frac{1}{6}\right)^{n-1}\end{align*}$
     ::=1=153(16n-1)
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  8. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{5} (1.05)^{n-1}\end{align*}$
     ::=1=1=15(1.05)-1
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  9. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{4}{7} \left(\frac{6}{7}\right)^{n-1}\end{align*}$
     ::Nn=147(67n-1)
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  10. $\begin{align*}\sum \limits_{n=1}^{\infty} 15 \left(\frac{11}{12}\right)^{n-1}\end{align*}$
      ::=1 =15(1112)n-1
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  11. $\begin{align*}\sum \limits_{n=1}^{\infty} 0.01 \left(\frac{3}{2}\right)^{n-1}\end{align*}$
      ::=1 0.01 (32)n-1
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  12. $\begin{align*}\sum \limits_{n=1}^{\infty} 100 \left(\frac{1}{5}\right)^{n-1}\end{align*}$
      ::Nn=1100( 150- 1)
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  13. $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{5}{4}\right)^{n-1}\end{align*}$
      ::=1 =12(54)-1
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  14. $\begin{align*}\sum \limits_{n=1}^{\infty} 2.5(0.85)^{n-1}\end{align*}$
      ::=1 =2.5(0.85)n-1
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  15. $\begin{align*}\sum \limits_{n=1}^{\infty} -3 \left(\frac{9}{16}\right)^{n-1}\end{align*}$
      ::=13(916)-1

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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 。