Course: 9.3 无限系列介绍

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无限系列介绍介绍

The rational number $\begin{align*}\frac{4}{9}\end{align*}$ can be written as $\begin{align*}0.44444 \ldots=0.\overline{4}\end{align*}$ . This repeating decimal can be written as an infinite series as follows:
::合理数字 49 可以写为 0.44444... =0. 4。这个重复的小数点可以写为无限序列, 如下:

\begin{aligned} \frac{4}{9} & = 0.4 + 0.04 + 0.004 + 0.0004 + \dots \\ = \frac{4}{10} + \frac{4}{100} + \frac{4}{1000} + \frac{4}{10, 000} + \dots \\ = \frac{4}{10} + \frac{4}{10^{2}} + \frac{4}{10^{3}} + \frac{4}{10^{4}} + \dots \end{aligned}

$\begin{align*}\frac{4}{9} &= 0.4 +0.04+0.004+0.0004+\ldots\\ &= \frac{4}{10}+\frac{4}{100}+\frac{4}{1000}+\frac{4}{10,000}+\ldots\\ &= \frac{4}{10}+\frac{4}{10^2}+\frac{4}{10^3}+\frac{4}{10^4}+\ldots\end{align*}$

Any repeating decimal can be written as an infinite series that represents a finite value, a rational value. Can you name the type of infinite series that can be used to represent the finite value of any repeating decimal? Can you write the above fraction $\begin{align*}\frac{4}{9}\end{align*}$ as an infinite series using summation notation?
::任何重复的小数点可以写成代表一定值、理性值的无限序列。您能够指定可以用来代表任何重复小数点的有限值的无限序列类型吗? 您能否用加号符号将以上第49个小数点写成一个无限序列?

Infinite Series
::无限系列

In the previous concepts, we looked at infinite sequences $\begin{align*}\{a_n\}\end{align*}$ and their characteristics of convergence or divergence. This concept extends the concept of an infinite to an infinite series, which is just the infinite sum of the sequence members. An infinite series is the sum $\begin{align*}S\end{align*}$ of the terms of an infinite sequence, $\begin{align*}u_1, u_2, u_3, u_4, \ldots\end{align*}$ , usually written as
::在先前的概念中,我们研究了无限序列 {an} 及其趋同或分歧的特性。这个概念将无限的概念延伸至无限序列的概念,这仅仅是序列成员无限的和。一个无限的序列是无限序列条件的S和,u1,u2,u3,u4...,通常写为:

\begin{aligned} S = u_{1} + u_{2} + u_{3} + u_{4} + \dots \end{aligned}

$\begin{align*}S=u_1+u_2+u_3+u_4+ \ldots\end{align*}$
::S=1+u2+u3+u4+...

or, in sigma notation as:
::或,以污名表示:

\begin{aligned} S = \sum_{k = 1}^{\infty} u_{k} . \end{aligned}

$\begin{align*}S=\sum\limits^{\infty}_{k=1} u_k.\end{align*}$
::Sk=1uk。 k=1uk。

(Read “the sum of the terms $\begin{align*}u_k\end{align*}$ for $\begin{align*}k\end{align*}$ equal to 1 to infinity.”)
:读 " k 等于 1 至无穷的 uk 术语之和 " 。 )

For practice, based on the given sequence, define the infinite series in terms of the sum of terms and using sigma notation:
::就实践而言,根据给定的顺序,用术语之和和和使用污名符号来界定无限序列:

1, 0.1, 0.01, 0.001, ...

$\begin{align*}\{a_n=n\}\end{align*}$ for $\begin{align*}n \ge 1\end{align*}$ .
::{an=n} n1 的 {an=n} for n1 {an=n} for n1_BAR_
1.5, -3.0, 6.0, -12.0, ...
11, 7, 3, -1, -4, -8, ...

The infinite sequence 1, 0.1, 0.01, 0.001, ... can have the following infinite series representations:
::无限序列1, 0.1, 0.01, 0.001, .可以有以下无限序列表示:

$\begin{align*}S=1+0.1+0.01+0.001+\ldots\end{align*}$
::S=1+0.1+0.01+0.001+...

$\begin{align*}S=\sum\limits^\infty_{k=1} \left(\frac{1}{10}\right)^{k-1}\end{align*}$
::Sk=1(110)k-1

The infinite sequence $\begin{align*}\{a_n=n\}\end{align*}$ can have the following infinite series representations:
::无限序列 {an=n} 可以包含以下无限序列表达式 :

$\begin{align*}S=0+1+2+3+4+\ldots\end{align*}$
::S=0+1+2+3+4+...

$\begin{align*}S=\sum\limits^\infty_{k=1} k\end{align*}$
::Sk=1k

The infinite sequence 1.5, -3.0, 6.0, -12.0, ... can have the following infinite series representations:
::无限序列1.5, -3.0, 6.0, -12.0,... 可以有以下无限序列表示:

$\begin{align*}S=1.5-3.0+6.0-12.0+\ldots\end{align*}$
::S=1.5-3.0+6.0-12.0+...

$\begin{align*}S=\sum\limits^\infty_{k=1}1.5 (-2)^{k-1}\end{align*}$
::Sk=11.5(--2)k-1

The infinite sequence 11, 7, 3, -1, -4, -8, ... can have the following infinite series representations:
::无限序列11, 7, 3, 3,1, 4,8,... 可以有以下无限序列表示 :

$\begin{align*}S=11+7+3-1-4-8 \ldots\end{align*}$
::S=11+7+3 -1 -4 -8...

$\begin{align*}S=\sum\limits^\infty_{k=1} [11-3(k-1)]\end{align*}$
::Sk=1[11-3(k-1)]

Since the individual members $\begin{align*}u_k\end{align*}$ of a sequence could be negative, zero, or positive, so can the individual terms of the corresponding infinite series. Because important properties of a series are based on whether the series terms are all positive, or mixed (negative, zero, or positive), it is important to be able to recognize a series as being in one of the following two categories:
::由于一个序列的单个成员uk可以是负的、零的或正的,相应的无限序列的单个术语也可以是正的。由于一个序列的重要属性基于是正的还是混合的(负的、零的或正的)系列术语,因此必须能够承认一个序列属于以下两类中的一类:

Positive (Non-negative) term series , or
:非负)正(非负)术语序列,或

Positive and negative term series (the simplest form is the alternating series).
::正面和负面术语系列(最简单的形式是交替序列)。

Two Categories of Infinite Series
Series type	Expanded Form	Sigma Notation	Parameters
Positive Term	$\begin{align}S=a_1+a_2+a_3+ \ldots +a_n+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1}a_n\end{align}$	$\begin{align}a_n > 0\end{align}$
Positive & Negative Term: ::正负时数 : Alternating ::互换	$\begin{align}S=a_1-a_2+a_3-\ldots+(-1)^{n-1}a_n+\ldots\end{align}$ ::S=1-a2+a3-...+(-1)n-1an+... or ::或 $\begin{align}S=-a_1+a_2-a_3+\ldots +(-1)^n a_n+\ldots\end{align}$ ::Sa1+a2 -a3+...+(- 1)+...	$\begin{align}S=\sum\limits^\infty_{n=1}(-1)^{n-1}a_n\end{align}$ ::Sn=1(-1-1-n-1an) or ::或 $\begin{align}S=\sum\limits^\infty_{n=1} (-1)^n a_n\end{align}$ ::Sn=1(- 1) nan	$\begin{align}a_n > 0\end{align}$

In addition, there are some important types of series that will appear frequently in subsequent concepts, and that you should learn to recognize. These are shown in the following table:
::此外,有些重要的系列将经常出现在以后的概念中,你应该学会认识。

Common Series Types
Series Type	Expanded Form	Sigma Notation	Parameters
Arithmetic	$\begin{align}S=t_1+(t_1+d)+(t_1+2d)+(t_1+3d)+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} [t_1+d(n-1)]\end{align}$	$\begin{align}t_1, d\end{align}$ are constants.
Geometric	$\begin{align}S=a+ar+ar^2+ar^3+\ldots ar^{n-1}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} ar^{n-1}\end{align}$	$\begin{align}a,r \end{align}$ are constants.
Harmonic	$\begin{align}S=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} \frac{1}{n}\end{align}$	---
$\begin{align}p\end{align}$ -Seies	$\begin{align}S=1+\frac{1}{2^p}+\frac{1}{3^p}+\ldots+\frac{1}{n^p}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} \frac{1}{n^p}\end{align}$	$\begin{align}p>0\end{align}$

Let's identify the characteristics of each of the following series using the type information above:
::让我们使用上述类型的信息,确定以下系列中每一系列的特性:

$\begin{align*}-4.3 -3.1 -1.9 -0.7 +\ldots\end{align*}$
$\begin{align*}\frac{1}{72}+\frac{1}{24}+\frac{1}{8}+\ldots\end{align*}$
$\begin{align*}1-0.6+0.06-0.006+\ldots\end{align*}$

Each term of the series $\begin{align*}-4.3 -3.1 -1.9 -0.7 +\ldots\end{align*}$ is 1.2 greater than the previous term. This means the series is an infinite, positive/negative term arithmetic with first term $\begin{align*}t_1=-4.3\end{align*}$ , and common difference $\begin{align*}d=1.3\end{align*}$ .
::系列-4.3-3.1-1.9-0.7+.中的每个术语比上一个术语大1.2。这意味着该序列是一个无限的、正/负的计算术语,第一个术语为t14.3,共同差 d=1.3。

After the first term of the series $\begin{align*}\frac{1}{72}+\frac{1}{24}+\frac{1}{8}+\ldots\end{align*}$ each term is 3 times greater than the previous term. This means the series is an infinite, positive term geometric with first term $\begin{align*}t_1=\frac{1}{72}\end{align*}$ , and common ratio $\begin{align*}r=3\end{align*}$ .
::在172+124+18+系列的第一个任期之后,每个任期比上一个任期大3倍。这意味着该系列是一个无限的、正数术语,第一任期为t1=172,共同比率为r=3。

After the first term of the series $\begin{align*}1-0.6+0.06-0.006+\ldots\end{align*}$ each term is $\begin{align*}-\frac{1}{10}\end{align*}$ times greater than the previous term. This means the series is an infinite, positive/negative (alternating) term geometric with first term $\begin{align*}t_1=1\end{align*}$ , and common ratio $\begin{align*}r=-\frac{1}{10}\end{align*}$ .
::1-0.6+0.06-0.006+系列的第一个任期之后,每个任期比上一个任期大110倍。这意味着该系列是一个无限的、正的/负的(交替的)术语几何,第一任期t1=1,和共同比率 r110。

Sequence of Partial Sums
::部分总和的顺序

As will be seen in the next concept, the sequence of finite (partial) sums $\begin{align*}\{s_n\}\end{align*}$ generated from the terms of any infinite series $\begin{align*}\sum\limits^\infty_{n=1} u_n\end{align*}$ tells a lot about the convergence (or divergence) properties of the infinite series. For the series $\begin{align*}\sum\limits^\infty_{n=1} u_n\end{align*}$ , the sequence members are
::从下一个概念中可以看到,从任何无限序列"n=1"中生成的有限(部分)总和 {sn} {sn} 的顺序,可以大量说明无限序列的趋同(或差异)特性。对于“n=1”系列,序列成员是:

\begin{aligned} s_{1} & = u_{1} \\ s_{2} & = u_{1} + u_{2} \\ s_{3} & = u_{1} + u_{2} + u_{3} \\ ⋮ & ⋮ \\ s_{n} & = u_{1} + u_{2} + u_{3} + \dots + u_{n} \end{aligned}

$\begin{align*}s_1 &= u_1\\ s_2 &= u_1+u_2\\ s_3 &= u_1+u_2+u_3\\ \vdots & \qquad \qquad \vdots\\ s_n &= u_1+u_2+u_3+\ldots+u_n\end{align*}$
::s1=u1s2=u1+u2s3=u1+u2+u3{sn=u1+u2+u3+...+un

The first sum is the first term of the series. The second sum is the sum of the first two terms of the series. The third term is the sum of the first three terms. Thus, the $\begin{align*}n\end{align*}$ th finite (partial) sum, $\begin{align*}s_n\end{align*}$ , is the sum of the first $\begin{align*}n\end{align*}$ terms of the infinite series: $\begin{align*}s_n=u_1+u_2+u_3+\ldots+u_n\end{align*}$ . The sequence of partial sums is defined as:
::第一个总和是该系列的第一个任期。第二个总和是该系列头两个任期的总和。第三个任期是前三个任期的总和。因此,nnn-nimited(部分)总和是无限系列的第一个n条件的总和: sn=u1+u2+u3+...+un。部分总和的顺序定义为:

A Partial Sum of an Infinite Series
::无限系列的部分总和

For an infinite series $\begin{align*}\sum\limits^\infty_{k=1} u_k\end{align*}$ , the $\begin{align*}n\end{align*}$ th partial sum , $\begin{align*}s_n\end{align*}$ , is the sum of the first $\begin{align*}n\end{align*}$ terms of the infinite series is:
::对于无限序列 k=1uk,nthpart 和, sn, 是无限序列第一个 n 条件的和 :

\begin{aligned} s_{n} = \sum_{k = 1}^{n} u_{k} . \end{aligned}

$\begin{align*}s_n=\sum\limits^n_{k=1}u_k.\end{align*}$
::{\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! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!

The sequence $\begin{align*}\{s_n\}\end{align*}$ formed from these sums is called the sequence of partial sums .
::由这些金额构成的顺序 {sn} 称为部分金额的顺序。

As you can see, the set of sums $\begin{align*}s_n\end{align*}$ form a sequence. The sequence is very important for the study of the related infinite series for it tells a lot about the infinite series.
::正如你所看到的, 一组总和 Sn 形成一个序列。序列对于相关无限序列的研究非常重要, 因为它能告诉很多关于无限序列的内容。

For practice, let's find the sequence of the first $\begin{align*}n\end{align*}$ of each infinite series:
::为了实践,让我们找到每个无限序列的第一个 n 序列的顺序 :

$\begin{align*}1+0.1+0.01+0.001+\ldots, n=5\end{align*}$ .
- This is a positive term geometric series. The sequence of the first five partial sums is: $\begin{align*}\{s_1, s_2, s_3, s_4, s_5\}\end{align*}$ is:
  ::这是正数数数序列。前五个部分总和的顺序是 {s1, s2, s3, s4, s5} :
  
  $\begin{aligned} s_{1} & = u_{1} = 1 \\ s_{2} & = u_{1} + u_{2} = 1 + 0.1 = 1.1 \\ s_{3} & = 1 + 0.1 + 0.01 = 1.11 \\ s_{4} & = 1 + 0.1 + 0.01 + 0.001 = 1.111 \\ s_{5} & = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.1111 \end{aligned}$
  $\begin{align*}s_1 &= u_1=1\\ s_2 &= u_1+u_2=1+0.1=1.1\\ s_3 &= 1+0.1+0.01=1.11\\ s_4 &= 1+0.1+0.01+0.001=1.111\\ s_5 &= 1+0.1+0.01+0.001+0.0001=1.1111\end{align*}$
  ::s1=u1=1s2=u1+u2=1+0.1=1+0.1=1.1s3=1+0.1+0.01=1.11s4=1+0.1+0.01+0.001=1.111=1+1.111=1+0.1+0.01+0.001+0.001=1.111
  
  For a geometric series with first term $\begin{align*}a\end{align*}$ and common ratio $\begin{align*}r\end{align*}$ , the $\begin{align*}n\end{align*}$ th partial sum $\begin{align*}S_n\end{align*}$ is computed from: $\begin{align*}S_n=a \frac{1-r^n}{1-r}\end{align*}$ .
  ::对于具有第一任期a和共同比率r的几何序列,从Sn=a1-rn1-r计算出第n部分和Sn:n=a1-rn1-r。
  
  ::这是一个正数数数序列。前五个部分总和的顺序是 {s1, s2, s3, s4, s5} : s1= u1=1s2=u1+u2=1+0.1=1x3=1+0.1x3=1+0.1+0.01=1.11s4=1+0.1+0.001+0.001=1.111s5=1+0.1+0.01+0.001+0.001+0.001=1.111; 对于第一个和共同比率为r的几部分数序列,则计算为nn=a1-rnn1-r。
::1+0.1+0.01+0.001+...,n=5. 这是正数几何序列。前五个部分总和的顺序是 {s1, s2, s3, s4, s5} : s1= u1=1s2=1s2=1+u2=1+0.1=1.11s3=1+0.1+0.01=1.11s4=1+0.1+0.01+0.001+1=1.111s5=1+0.1+0.001+0.001+0.001+0.001=1.1111。对于第一个和共同比率为r的几何序列,第五个部分总和Sn的顺序是: Sn=a1-rn1-r。

$\begin{align*}S=\sum\limits^\infty_{k=1} k, n=10\end{align*}$ .
- Perhaps you recall from a previous concept that $\begin{align*}S_n=\sum\limits^n_{k=1} k =\frac{n(n+1)}{2}\end{align*}$ . Using this shortcut, allows the determination that the sequence of the first ten partial sums $\begin{align*}\{s_1, s_2, s_3, \ldots, s_{10}\}\end{align*}$ is:
  ::也许您还记得之前的一个概念, Snák=1nk=n(n+1) 2。使用此快捷键, 可以确定前十部分金额 {s1, s2, s3,..., s10} 的顺序是 :
  
  $\begin{aligned} s_{1} = \frac{1 (1 + 1)}{2} = 1 & s_{6} = 21 \\ s_{2} = \frac{2 (2 + 1)}{2} = 3 & s_{7} = 28 \\ s_{3} = 6 & s_{8} = 36 \\ s_{4} = 10 & s_{9} = 45 \\ s_{5} = 15 & s_{10} = 55 \end{aligned}$
  $\begin{align*}& s_1=\frac{1(1+1)}{2}=1 & s_6=21\\ & s_2 =\frac{2(2+1)}{2}=3 & s_7=28\\ & s_3=6 & s_8=36\\ & s_4=10 & s_9=45\\ & s_5=15 & s_{10}=55\end{align*}$
  ::s1=1( 1+1) 2=1s6=21s2=2( 2+1) 2=3s7=28s3=6s8=36s4=10s9=45s5=15s10=55
  
  ::您也许还记得上一个概念中, Snk=1nk=n( n+1) 2。使用此快捷键, 可以确定前十部分总和 {s1, s2, s3,..., s10} 的顺序是: s1=1( 1+1) 2=1s6=1s6=21s2=2( 2+1) 22=3s7=28s3=6s8=6s8=36s4=10s9=45s5=15s10=55
::Sk=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

$\begin{align*}S=\sum\limits^{\infty}_{k=1} 1.5 (-2)^{k-1}, n=10\end{align*}$
- This is an alternating geometric series with $\begin{align*}a=1.5\end{align*}$ and $\begin{align*}r=-2\end{align*}$ . The sequence of the first ten partial sums $\begin{align*}\{s_1, s_2, s_3, \ldots s_{10}\}\end{align*}$ is:
  ::这是一个交替的几何序列, a=1.5 和 r2 。前十部分总和 {s1, s2, s3,...s10} 的顺序是 :
::Sk=11.5(- 2,k- 1,n=10) 这是一个交替的几何序列, a=1.5 和 r=2 。前十部分总和 {s1, s2, s3,...s10} 的顺序是 :

\begin{aligned} s_{1} = \sum_{k = 1}^{1} 1.5 (- 2)^{1 - 1} = 1.5 & s_{6} = - 31.5 \\ s_{2} = \sum_{k = 1}^{2} 1.5 (- 2)^{2 - 1} = 1.5 - 3 = - 1.5 & s_{7} = 64.5 \\ s_{3} = \sum_{k = 1}^{3} 1.5 (- 2)^{k - 1} = 4.5 & s_{8} = - 127.5 \\ s_{4} = \sum_{k = 1}^{4} 1.5 (- 2)^{k - 1} = - 7.5 & s_{9} = 256.5 \\ s_{5} = \sum_{k = 1}^{5} 1.5 (- 2)^{k - 1} = 16.5 & s_{10} = - 511.5 \end{aligned}

$\begin{align*}& s_1=\sum\limits^1_{k=1} 1.5(-2)^{1-1}=1.5 && s_6=-31.5\\ & s_2=\sum\limits^2_{k=1} 1.5(-2)^{2-1}=1.5-3=-1.5 && s_7=64.5\\ & s_3=\sum\limits^3_{k=1} 1.5(-2)^{k-1}=4.5 && s_8=-127.5\\ & s_4=\sum\limits^4_{k=1} 1.5(-2)^{k-1}=-7.5 && s_9=256.5\\ & s_5=\sum\limits^5_{k=1} 1.5(-2)^{k-1}=16.5 && s_{10}=-511.5\end{align*}$
::=111.55(-2)1-1=1.5s6-31.5s2-k=111.55(-2)2-2-1=1.5-31.5s7=64.5s3}k=131.5(-2k-1)=4.5s8-127.5s4k=141.5(-2k-1)-7.5s9=256.5s5}k=151.5(2k-1)=151.5(2k-1)=16.5s10_511.5

Examples
::实例

Example 1
::例1

Earlier, you were asked to name the type of infinite series that can be used to represent the finite value of any repeating decimal. Can you write the fraction $\begin{align*}\frac{4}{9}\end{align*}$ as an infinite series using summation notation?
::早些时候, 您被要求指定可以用来代表任何重复的小数点的有限值的无限序列类型。您能否用加号符号将第49项作为无限序列写入 ?

The geometric series is the name of the infinite series that can be used to represent the finite value of any repeating decimal. It has the form $\begin{align*}S=\sum\limits^\infty_{n=1} ar^{n-1}\end{align*}$ . Since the fraction $\begin{align*}\frac{4}{9}\end{align*}$ can be written as $\begin{align*}\frac{4}{9}=\frac{4}{10}+\frac{4}{10^2}+\frac{4}{10^3}+\frac{4}{10^4}+\ldots\end{align*}$ , the geometric series is written as $\begin{align*}\frac{4}{9}=\sum\limits^\infty_{n=1} \frac{4}{10} \left(\frac{1}{10}\right)^{n-1}=\sum\limits^\infty_{n=1} \frac{2}{5} \left(\frac{1}{10}\right)^{n-1}\end{align*}$ .
::几何序列是可用于表示任何重复小数点的有限值的无限序列的名称。它的表Sn=1arn-1。由于第49项可以写为 49= 410+4102+4103+4104+..., 因此几何序列的写法是 49n=1410(110)n- 1n=125(110)n-1。

Example 2
::例2

For the series $\begin{align*}7, -\frac{21}{2}, \frac{63}{3}, -\frac{189}{8}, \ldots\end{align*}$ , describe the stated series using as many of the following descriptors that apply:
::在7-212,633,-1898系列中,使用下列适用的说明性说明来描述所述系列:

Finite or Infinite
::有限或无限

Positive term or positive and negative term
::积极或积极和消极术语

Arithmetic, geometric, harmonic, $\begin{align*}p\end{align*}$ -series, or unknown type
::亚学、几何、口音、p系列或不明类型

The series continues indefinitely and alternates between positive and negative terms: it is an infinite . Also the ratios between the 2 ^nd and 1 ^st terms, the 3 ^rd and 2 ^nd terms, etc. are the same, $\begin{align*}-\frac{3}{2}\end{align*}$ . This is a geometric series with common ratio $\begin{align*}r=-\frac{3}{2}\end{align*}$ and $\begin{align*}a=7\end{align*}$
::该系列将无限期延续,并在正值和负值之间交替:这是一个无限的。同样,第二和第一期、第三和第二期等之间的比率是相同的,-32。这是一个具有共同比率的几何序列,32和a=7。

Example 3
::例3

For the same series $\begin{align*}7, -\frac{21}{2}, \frac{63}{3}, -\frac{189}{8}, \ldots\end{align*}$ , write the series using sigma notation.
::同样的系列7 - 212,633, - 1898, 使用Sigma 符号来写。

The series can be written in sigma notation as $\begin{align*}S=\sum\limits^\infty_{n=1} 7 \left(-\frac{3}{2}\right)^{n-1}\end{align*}$ .
::该系列可以以 sigma 符号 as Sn= 1\\\7(- 32)n- 1 来写成。

Example 4
::例4

For the same series $\begin{align*}7, -\frac{21}{2}, \frac{63}{3}, -\frac{189}{8}, \ldots\end{align*}$ , compute the 10 ^th term of the series and the 3 ^rd partial sum.
::同一系列7-212,633,-1898,计算该系列的第10期和第3部分总和。

For a geometric series, the $\begin{align*}n\end{align*}$ th term is given by $\begin{align*}t_n=ar^{n-1}:\end{align*}$
::对于几何序列, nth 术语由 tn=arn- 1 给出 :

\begin{aligned} t_{10} = a r^{10 - 1} = 7 {(- \frac{3}{2})}^{9} = - \frac{137781}{512} . \end{aligned}

$\begin{align*}t_{10}=ar^{10-1}=7 \left(-\frac{3}{2}\right)^9=-\frac{137781}{512}.\end{align*}$
::t10=ar10-1=7(-32)9137781512。

For a geometric series, the $\begin{align*}n\end{align*}$ th partial sum is given by $\begin{align*}S_n=a \frac{1-r^n}{1-r}\end{align*}$ :
::对于几何序列,Nn=a1-rn1-r给出了 n部分和:

\begin{aligned} S_{3} = 7 \frac{1 - {(- \frac{3}{2})}^{3}}{1 - (- \frac{3}{2})} = \frac{49}{4} . \end{aligned}

$\begin{align*}S_3=7 \frac{1-\left(-\frac{3}{2}\right)^3}{1-\left(-\frac{3}{2}\right)}=\frac{49}{4}.\end{align*}$
::S3=71-(-32-32)31-(-32-32)=494。

Review
::回顾

For #1-5, express the number as an infinite series in sigma notation form.
::对于 # 1 5, 表示数字是一个无限的序列, 以污名表示。

$\begin{align*}\frac{3}{9}\end{align*}$
$\begin{align*}\frac{1}{11}\end{align*}$
$\begin{align*}0.\overline{037}\end{align*}$
$\begin{align*}0.\overline{1573}\end{align*}$
0.12012001200012...

For #6-9, identify the type of infinite series with as much detail as possible (positive or positive/negative term, arithmetic with first term and common difference, geometric, etc.).
::对于#6-9,用尽可能详细的方式(正或正/负术语、算术与第一个术语和共同差异、几何等)标明无限序列的类型。

$\begin{align*}-7-3+1+5+9+\ldots\end{align*}$

$\begin{align*}\sum\limits^\infty_{k=1} \left(\frac{3}{5}\right)^{k-1}\end{align*}$
::k=1( 35)k- 1

$\begin{align*}\sum\limits^\infty_{k=1} \frac{k^3}{k^3-5}\end{align*}$
::k=1k3k3-5

$\begin{align*}\sum\limits^\infty_{k=1} \frac{4^{k+2}}{9^{k-1}}\end{align*}$
::k=14k+29k-1

For #10-15, identify the series characteristics and find the first three partial sums, $\begin{align*}s_1, s_2, s_3\end{align*}$ :
::对于#10-15, 标明序列特性, 并找到前三个部分总和, s1, s2, s3:

$\begin{align*}\sum\limits^\infty_{k=1} \frac{k+2}{5^{k-1}}\end{align*}$
::k=1k+25k-1

$\begin{align*}\sum\limits^{\infty}_{n=1} 5 \left(-\frac{1}{2}\right)^{n-1}\end{align*}$
::=1 =5 (- 12) -1

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

$\begin{align*}\sum\limits^{\infty}_{i=1} i^3\end{align*}$
::i=1i3

$\begin{align*}\sum\limits^{\infty}_{n=1} \frac{7}{n}\end{align*}$
::=1 =7n =1 =7n

$\begin{align*}\sum\limits^{\infty}_{n=1} \frac{30}{n^2}\end{align*}$
::=1=1 =30n2

Review (Answers)
::回顾(答复)

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。

Series Type	Expanded Form	Sigma Notation	Parameters
Arithmetic	$\begin{align}S=t_1+(t_1+d)+(t_1+2d)+(t_1+3d)+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} [t_1+d(n-1)]\end{align}$	$\begin{align}t_1, d\end{align}$ are constants.
Geometric	$\begin{align}S=a+ar+ar^2+ar^3+\ldots ar^{n-1}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} ar^{n-1}\end{align}$	$\begin{align}a,r \end{align}$ are constants.
Harmonic	$\begin{align}S=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} \frac{1}{n}\end{align}$	---
$\begin{align}p\end{align}$ -Seies	$\begin{align}S=1+\frac{1}{2^p}+\frac{1}{3^p}+\ldots+\frac{1}{n^p}+\ldots\end{align}$	$\begin{align}S=\sum\limits^\infty_{n=1} \frac{1}{n^p}\end{align}$	$\begin{align}p>0\end{align}$

Section outline

General

无限系列介绍介绍

Infinite Series ::无限系列

Sequence of Partial Sums ::部分总和的顺序

A Partial Sum of an Infinite Series ::无限系列的部分总和

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)