9.4 确定无限系列的趋同或分歧

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• 选择节确定无限系列的趋同或分歧

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  确定无限系列的趋同或分歧

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  Suppose you stand exactly 8 feet from a wall, and begin taking steps toward the wall by halving the distance remaining with each step. Can you express mathematically, using sigma notation, how far you have gone after each step? What kind of series is it? What does the series tell you about the number of steps it would take to actually get to the wall?
  ::假设你站在离墙8英尺处,开始向墙走一步,将每一步所剩距离减半。你能用数学表达方式,用刻度符号表示,每步后你走多远?是哪一系列?系列是什么?系列中有多少步骤可以真正到达墙上?

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  Convergence and Divergence of an Infinite Series
  ::无限系列的趋同和分歧

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  Just as with sequences, we can talk about convergence and divergence of infinite series. It turns out that the convergence or divergence of an infinite series depends on the convergence or divergence of the .
  ::与序列一样,我们可以谈论无限系列的趋同和分歧。 事实证明,无限系列的趋同或分歧取决于这些序列的趋同或分歧。

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  Let $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ be an infinite series, and let $\begin{align*}\{s_n\}\end{align*}$  be the sequence of partial sums for the series:
  ::@k=1uk 是一个无限的序列, 让{sn} 成为序列部分总和的序列 :

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  • If $\begin{align*}\lim\limits_{n \to \infty} s_n=S\end{align*}$ , where $\begin{align*}S\end{align*}$  is a real number, then the infinite series converges and $\begin{align*}\sum\limits_{k=1}^\infty u_k=S\end{align*}$ .
    ::如果 limnsn=S, 其中的 S 是真实数字, 那么无限序列会相交 。 @k=1uk=S 。
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  • If $\begin{align*}\lim\limits_{n \to \infty} s_n\end{align*}$  does not have a finite limit, then the infinite series diverges .
    ::如果limnsn没有限制,那么无限的序列就会不同。

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  If we can determine whether the limit of the partial sum of an infinite series is finite, then we know the infinite series converges.
  ::如果我们能确定无限序列部分总和的限度是否是有限的,那么我们就知道无限序列的趋同。

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  Based on the above definitions of the convergence and divergence of an infinite series, the table below summarizes convergence/divergence for some of the common and recognizable infinite series presented in the previous concept. You should learn to recognize these series forms.
  ::根据上述关于无限系列的趋同和分歧的定义,下表总结了前一个概念提出的一些共同和可识别的无限系列的趋同/引力。你应该学会认识这些系列的形式。

  Convergence Properties of Common Series Types

  播放段落 Series Type ::系列系列类型	Sigma Notation	播放段落 Converges if ::组合	播放段落 Diverges If ::潜水点如果
  Arithmetic	$\begin{align*}S=\sum\limits_{n=1}^\infty [t_1+d(n-1)]\end{align*}$	Never	Always
  Geometric	$\begin{align*}S=\sum\limits_{n=1}^\infty ar^{n-1}\end{align*}$	$\begin{align*}|r|<1 \end{align*}$ with $\begin{align*}S=\frac{a}{1-r}\end{align*}$	$\begin{align*}|r| \ge 1\end{align*}$
  Harmonic	$\begin{align*}S=\sum\limits_{n=1}^\infty \frac{1}{n}\end{align*}$	Never	Always
  $\begin{align*}p\end{align*}$ -Series	$\begin{align*}S=\sum\limits_{n=1}^\infty \frac{1}{n^p}\end{align*}$	$\begin{align*}p>1\end{align*}$	$\begin{align*}p \le 1\end{align*}$

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  Let us apply the definition for convergence and divergence to some series types presented in the previous concept.
  ::让我们将趋同和分歧的定义适用于前一个概念提出的一些系列。

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  Determine whether the following infinite series converges or diverges: $\begin{align*}S=-100-95-90-85+ \ldots\end{align*}$
  ::确定以下无限序列是否相近或相异: S100-95-90-85+...

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  The infinite series $\begin{align*}S=-100-95-90-85+ \ldots\end{align*}$  can be written in sigma notation as $\begin{align*}S=\sum\limits_{k=1}^\infty [-100+5(k-1)]\end{align*}$ .
  ::无限序列 S100-95-90-85+...可以用Sk=1[-100+5(k-1)]的西格玛符号书写。

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  This series is an arithmetic series with  $\begin{align*}t_1=-100\end{align*}$ and $\begin{align*}d=5\end{align*}$ .
  ::这是一个计算序列, T1100 和 d=5 。

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  The $\begin{align*}n\end{align*}$ th partial sum, $\begin{align*}S_n\end{align*}$ , of an arithmetic series is given by
  ::计算序列的n部分总和(Sn)由下列人士提供:

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  $$\begin{align*}S_n=\frac{n}{2} \left[t_1 + t_n\right] = \frac{n}{2} \left[2t_1 + (n-1)d\right].\end{align*}$$
  ::Sn=n2[t1+tn]=n2[2t1+(n-1)d]。

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  Using the definition of series convergence/divergence by evaluating the limit of the $\begin{align*}n\end{align*}$ th partial sum yields:
  ::使用序列趋同/共振的定义,通过评价nth部分总和产量的限度:

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  $$\begin{align*}\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{n}{2} \left[2(-100) + (n-1)5\right] = \lim_{n \to \infty} \frac{n}{2} \left[-200\right] + \lim_{n \to \infty} \frac{n}{2} \left[(n-1)5 \right] = \infty.\end{align*}$$
  ::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

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  The infinite arithmetic series is divergent. This is true for all infinite arithmetic series!
  ::无限算术序列是不同的。所有无限算术序列都是如此!

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  Let’s look at another example.
  ::让我们再看看另一个例子。

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  Does the infinite series $\begin{align*}1+0.1+0.01+0.001+ \ldots \end{align*}$   converge or diverge ?
  ::无限序列 1+0. 1+0.01+0.001+... 趋同还是相左?

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  This is a geometric series with $\begin{align*}a=1\end{align*}$  and $\begin{align*}r=\frac{1}{10}=0.1\end{align*}$ .
  ::这是一个几何序列, a=1, r=110=0.1。

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  Recall from concepts in Analysis, that the $\begin{align*}n\end{align*}$ th partial sum, $\begin{align*}S_n\end{align*}$ , of a geometric series is given by
  ::回顾《分析分析》中的概念,即几几何序列的n部分和Sn由

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  S n = a 1 − r n 1 − r .

  $$\begin{align*}S_n=a \frac{1-r^n}{1-r}.\end{align*}$$
  ::Sn=a1 -rn1 -r.

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  Using the definition of series convergence/divergence by evaluating the limit of the $\begin{align*}n\end{align*}$ th partial sum yields:
  ::使用序列趋同/共振的定义,通过评价nth部分总和产量的限度:

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  $$\begin{align*}\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[a \frac{1-r^n}{1-r}\right] = \lim_{n \to \infty} \left[\frac{1-(0.1)^n}{1-0.1}\right] = \lim_{n \to \infty} \left[\frac{1}{0.9}\right] - \lim_{n \to \infty} \left[\frac{1-(0.1)^n}{0.9}\right] = \frac{10}{9}.\end{align*}$$
  ::[1-(0.1)n1--0.1]=(10.9)-(9)-(10.9)-[1-(0.1)n0.9]=109。

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  This is the result that would be obtained from the table above, since $\begin{align*}|r| < 1\end{align*}$ , the series converges to the value

  $$\begin{align*}S=\frac{a}{1-r}=\frac{1}{1-0.1}=\frac{10}{9}.\end{align*}$$
  ::这是从上表得出的结果,因为r1,该序列的数值是S=a1-r=11-0.1=109。

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  This solution can also be shown by the following process.
  ::这一解决办法也可以通过以下进程来证明。

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  Write the infinite series  $\begin{align*}1+0.1+0.01+0.001+ \ldots\end{align*}$ as an infinite series of fractions:
  ::写入无限序列 1+0. 1+0.01+0.001+...

  $$\begin{align*}1 + \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^ 3} + \cdots\end{align*}$$

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  The $\begin{align*}n\end{align*}$ th partial sum is: $\begin{align*}S_n=1 + \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^ 3} + \cdots + \frac{1}{10^{n-1}}\end{align*}$ .
  ::第n部分为:Sn=1+110+1102+1103+1103+110n-1。

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  Multiply both sides of the equation by $\begin{align*}\frac{1}{10}\end{align*}$ :
  ::乘以方程的两边乘以 110 :

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  $$\begin{align*}\frac{1}{10} s_n &= \frac{1}{10} \left(1+ \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \ldots + \frac{1}{10^{n-1}}\right) \\ &= \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + \ldots + \frac{1}{10^n}\end{align*}$$
  ::110sn=110(1+110+1102+1103+...+110n-1)=110+1102+1103+11004+...+110n

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  Now we have two equations:
  ::现在我们有两个方程式:

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  $$\begin{align*}s_n &=1+ \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \ldots + \frac{1}{10^{n-1}} \\ \frac{1}{10}s_n &= \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + \ldots + \frac{1}{10^n}\end{align*}$$
  ::sn=1+110+1102+1103+...+110n-1110sn=110+1102+1103+1104+...+110n

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  Subtract the bottom equation from the top equation to cancel terms and simplifying:
  ::从顶方程中减去底方程,取消条件和简化:

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  $$\begin{align*}s_n - \frac{1}{10} s_n &= 1- \frac{1}{10^n} \\ \frac{9}{10}s_n &= 1- \frac{1}{10^n}\end{align*}$$
  ::sn-110sn=1-110n910sn=1-110nn

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  Solve for $\begin{align*}s_n\end{align*}$  by multiplying both sides of the last equation by $\begin{align*}\frac{10}{9}\end{align*}$ :
  ::将最后一个方程的两边乘以109 来解决鼻鼻问题:

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  $$\begin{align*}s_n=\frac{10}{9} \left(1- \frac{1}{10^n}\right)\end{align*}$$
  ::sn=109(1-1-1-1100n)

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  Now we find the limit of both sides:
  ::现在,我们发现双方的界限是:

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  $$\begin{align*}\lim_{n \to +\infty} s_n &=\lim_{n \to \infty} \frac{10}{9} \left(1- \frac{1}{10^n}\right) \\ &=\lim_{n \to \infty} \frac{10}{9} - \lim_{n \to +\infty} \frac{10}{9} \left(\frac{1}{10^n}\right) \\ &=\frac{10}{9} - 0 = \frac{10}{9} \\ &= \frac{10}{9} \end{align*}$$
  ::=109 -0=109=109=109 =109=109=109=109=109=109=109=109=109=109=109=109=109=109=109=109=109=109=109

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  The sum of the infinite series is $\begin{align*}\frac{10}{9}\end{align*}$ , and so the series converges.
  ::无限序列的总和是109, 所以序列会汇合。

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  Determining convergence by using the limit of the sequence of partial sums is not always feasible or practical. This is the case with both the Harmonic and $\begin{align*}p\end{align*}$ -series, where there is no equation for the $\begin{align*}n\end{align*}$ th partial sum so that computing the limit is not possible. Other methods or tests to show convergence or divergence of these two, and other, series, must be used.
  ::通过使用部分金额序列的界限来确定趋同性并不总是可行或实际的,在调和和p系列中,情况都是如此,因为N部分和p系列没有等式,因此计算限额是不可能的,必须使用其他方法或试验来显示这两个系列和其他系列的趋同性或差异性。

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  If convergence/divergence cannot be shown by taking the limit of a partial sum expression, applying the following test to the $\begin{align*}n\end{align*}$ th partial sum term may show if a series is divergent.
  ::如果通过采用部分总和表达式的限度无法表明趋同/共性,那么,对n-部分总和适用下列检验标准,可以表明某一系列的差别。

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  The $\begin{align*}n\end{align*}$ th-Term Test for Series Divergence
  ::系列差异的 nth 期限测试

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  For an infinite series $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ , if $\begin{align*}\lim\limits_{k \to +\infty}u_k \ne 0\end{align*}$ or $\begin{align*}\lim\limits_{k \to +\infty}u_k \end{align*}$ does not exist , then the infinite series $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ diverges.
  ::对于无限序列 *k=1uk,如果limkuk=0或limkuk不存在,那么无限序列 *k=1uk的差异。

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  Note: If $\begin{align*}\lim\limits_{k \to +\infty}u_k=0\end{align*}$ , we cannot conclude anything about convergence or divergence.
  ::注:如果limkáuk=0,我们无法得出关于趋同或分歧的任何结论。

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  Determine whether the infinite series  $\begin{align*}S=\sum\limits_{n=1}^\infty \frac{1}{n^{-3}}\end{align*}$ converges or diverges.
  ::确定无限序列Sn=11n-3是否相近或不同。

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  This is a series of the form $\begin{align*}S=\sum\limits_{n=1}^\infty \frac{1}{n^p}\end{align*}$ , i.e., a $\begin{align*}p\end{align*}$ -series, with $\begin{align*}p=-3\end{align*}$ .
  ::这是表Sn=11np的系列, 即p系列, p3。

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  It is not tractable to obtain a closed for expression for the $\begin{align*}n\end{align*}$ th partial sum, which means we cannot apply the limit approach to determine convergence or divergence. However, the $\begin{align*}n\end{align*}$ th term test can be applied to test for divergence.
  ::以n部分总和表示的封闭式封闭式封闭式表达方式是行不通的,这意味着我们无法应用限制方法确定趋同或分歧。 但是, n术语测试可用于测试差异。

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  We have
  ::我们有

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  $$\begin{align*}\lim_{k \to +\infty} u_k = \lim_{k \to +\infty} \frac{1}{k^p} = \lim_{k \to +\infty} \frac{1}{k^{-3}} = \lim_{k \to +\infty} k^3 = \infty.\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦!

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  The series $\begin{align*}S=\sum\limits_{n=1}^\infty \frac{1}{n^{-3}}\end{align*}$  therefore diverges.
  ::因此,Sn=11n-3系列各有不同。

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  In general, any $\begin{align*}p\end{align*}$ -series diverges when $\begin{align*}p<0\end{align*}$ . Later concepts will give tests to show $\begin{align*}p\end{align*}$ -series convergence/divergence for other values of $\begin{align*}p\end{align*}$ .
  ::一般说来,任何p系列在p<0时都会有差异。后来的概念将进行测试,以显示p系列的趋同/引力对p的其他值的趋同/引力。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to express mathematically how far you have walked after each step if you begin taking steps toward a wall that is 8 feet away by halving the distance remaining with each step. What kind of series is it? What does the series tell you about the number of steps it would take to actually get to the wall?
  ::早些时候,有人要求你用数学来表达你走过每一步之后所走过的路程,如果你开始采取步骤,向距离8英尺远的墙走去,将每步所剩距离减半。这是哪一系列?该系列告诉你多少步步才能真正到达墙上?

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  The distance traveled, $\begin{align*}D_N\end{align*}$ , after $\begin{align*}N\end{align*}$  steps would look like:
  ::距离距离,DN,在N步骤之后, 看起来像:

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  $$\begin{align*}D_N=8 \left(\frac{1}{2}\right) + 8 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + 8 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \cdots = 8 \left(\frac{1}{2}\right) \left(1+ \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \cdots \left(\frac{1}{2}\right)^{N-1}\right) = \sum\limits_{k=1}^N 4 \left(\frac{1}{2}\right)^{k-1}\end{align*}$$
  ::DN=8( 12) +8( 12) (12) +8( 12) (12) +8( 12) (12) 8( 12) (1+12) +( 12) +( 12) +( 12) +( 12) +( 12) +( 12) +( 12) 2( (12) +( 12) 2( (12) N- 1) = Nk=14( 12) k- 1

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  This is a finite geometric series with  $\begin{align*}a=4\end{align*}$ feet and common ratio $\begin{align*}r=\frac{1}{2}\end{align*}$ . The $\begin{align*}n\end{align*}$ th partial sum of this geometric series is  $\begin{align*}S_N=a \frac{1-r^N}{1-r}=4 \frac{1 - \left(\frac{1}{2}\right)^N}{1- \frac{1}{2}}=8 \left[1- \left(\frac{1}{2}\right)^N\right]\end{align*}$ . Looks like you’ll never get to the wall! But it’s good to know that in the limit as $\begin{align*}N \rightarrow \infty\end{align*}$ , you do. Got time on your hands?
  ::这是一个限定的几何序列, 有 = 4 英尺, 通用比率 r= 12 。 这个几何序列的 nth 部分总和是 SN=a1 -rN1 -r=41 -( 12) N1 - 12=8 [ 1 - ( 12N ) 。 看起来你永远也到不了墙面 。 但是, 很高兴知道, 在极限内, 你这样做了。 有时间吗 ?

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  For each of the following examples, determine if the series converges or diverges.
  ::对于以下每一例,确定该系列的趋同或相异。

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  Example 2
  ::例2

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

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  $\begin{align*}\sum\limits_{k=1}^\infty \left(\frac{5}{6^{k-1}}\right) = \sum\limits_{k=1}^\infty 5 \left[\frac{1}{6}\right]^{k-1}\end{align*}$ . This is a geometric series with $\begin{align*}a=5\end{align*}$ and $\begin{align*}r=\frac{1}{6}\end{align*}$ . Because $\begin{align*}|r|<1\end{align*}$ , the series converges. It converges to the value $\begin{align*}S=\frac{a}{1-r}=\frac{5}{\frac{5}{6}}=6\end{align*}$ .
  ::k= 1 (56k- 1)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8\\\\\\\\\\\\8\\\\\\\\\\\几何数组。这是一个几何序列, 以a= 5andr=16。 因为\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Example 3
  ::例3

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  $\begin{align*}\sum\limits_{k=1}^\infty \frac{k}{k+5}\end{align*}$
  ::*k=1kk+5 *k=1kk+5

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  The series $\begin{align*}\sum\limits_{k=1}^\infty \frac{k}{k+5}\end{align*}$  is difficult to evaluate using the limit of the partial sum. Use the  $\begin{align*}n\end{align*}$ th-Term Test to determine if the series diverges.
  ::Qk=1kk+5序列很难使用部分总和的限度进行评估。 使用 n- 期测试来确定序列是否有差异 。

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  $$\begin{align*}\lim_{k \to +\infty} \frac{k}{k+5} &= \lim_{k \to +\infty} \frac{\frac{k}{k} }{\frac{k+5}{k}} \\ &= \lim_{k \to +\infty} \frac{1}{1+ \frac{5}{k}} \\ &= \frac{\lim\limits_{k \to +\infty} 1}{\lim\limits_{k \to +\infty} \left(1+\frac{5}{k}\right)} \\ &= 1\end{align*}$$
  ::kk+5=limkkkk+5k+5k=limk@5k+5k=11+5k=limk@1lik+5k=1

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  Because $\begin{align*}\lim\limits_{k \to +\infty} \frac{k}{k+5} \ne 0\end{align*}$ , the series $\begin{align*}\sum\limits_{k=1}^\infty \frac{k}{k+5}\end{align*}$ diverges.
  ::因为 limkkkk+5QQ0, 系列 *k=1kk+5的差异。

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  Example 4
  ::例4

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  $\begin{align*}\sum\limits_{k=1}^\infty \frac{8}{k-3}\end{align*}$
  ::kk=18k -3 k=18k -3

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  The series $\begin{align*}\sum\limits_{k=1}^\infty \frac{8}{k-3}\end{align*}$ is difficult to evaluate using the limit of the partial sum. Use the  $\begin{align*}n\end{align*}$ th- Term Test to determine if the series diverges:
  ::QQk=18k-3序列很难使用部分总和的限度进行评估。 使用 n- 期测试来确定序列是否有差异 :

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  $$\begin{align*}\lim_{k \to +\infty} \frac{8}{k-3} = \lim_{k \to + \infty} \frac{\frac{8}{k}}{\left ( \frac{1-3}{k} \right ) } = 0.\end{align*}$$
  ::立方公尺8k-3=立方公尺8k(1-3k)=0。

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  Since the limit is 0, we cannot make a conclusion about convergence or divergence.
  ::由于限制是0,我们无法就趋同或分歧得出结论。

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  Review
  ::回顾

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  For #1-13, determine if the infinite series converges or diverges. If the  series converges, find its sum.
  ::对于 # 1- 13, 确定无限序列是否趋同或相异。 如果该序列趋同, 请找到其总和 。

  1. $\begin{align*}3 + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots\end{align*}$
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  2. $\begin{align*}\sum\limits_{k=1}^\infty \left(\frac{3}{5}\right)^{k-1}\end{align*}$
     ::k=1 (35)k- 1
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  3. $\begin{align*}\sum\limits_{k=1}^{+\infty} \left(-\frac{2}{3}\right)^{k-1}\end{align*}$
     ::*k=1(- 23k- 1)
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  4. $\begin{align*}\sum\limits_{k=1}^\infty \frac{k^3}{k^3-5}\end{align*}$
     ::k=1k3k3 -5
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  5. $\begin{align*}\sum\limits_{k=1}^\infty \frac{4^{k+2}}{9^{k-1}}\end{align*}$
     ::k=14k+29k-1
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  6. $\begin{align*}7 + \frac{7}{8} + \frac{7}{8^2} + \frac{7}{8^3} + \cdots + \frac{7}{8^{i-1}} + \cdots\end{align*}$
     ::7+78+782+783+783+78i-1{___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  7. $\begin{align*}\sum\limits_{k=1}^{+\infty} 9^{k-1}\end{align*}$
     ::kk=19k-1 k=19k-1
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  8. $\begin{align*}-\frac{3}{4} + \frac{3}{4^2} - \frac{3}{4^3} + \cdots + \frac{3(-1)^k}{4^k} + \cdots\end{align*}$
     ::-34+342-3433(-1)k4k
  9. $\begin{align*}3+2+ \frac{4}{3} + \frac{8}{9} + \cdots\end{align*}$
  10. $\begin{align*}-2+ \frac{5}{2} - \frac{25}{8} + \frac{125}{32} + \cdots\end{align*}$
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  11. $\begin{align*}\sum\limits_{n=1}^\infty \frac{n}{n+7}\end{align*}$
      ::*n=1nn+7
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  12. $\begin{align*}\sum\limits_{n=1}^\infty \frac{n+1}{2n-3}\end{align*}$
      ::*n=1n+12n-3
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  13. $\begin{align*}\sum\limits_{n=1}^\infty \sqrt[n]{3}\end{align*}$
      ::=1 =1 =3 =1 =1 =1 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 3 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 3
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  14. For what values of $\begin{align*}x \end{align*}$ does the series $\begin{align*}\sum\limits_{k=1}^\infty 3 \sin(x)^{k-1}\end{align*}$ converge?
      ::对于 x 的值, kk=13sin( x) k- 1 序列会合并到什么值 ?
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  15. Prove that $\begin{align*}0.99999999 \ldots = 1\end{align*}$ .
      
      ::证明那0.9999999999... =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 。