Section: 阳性系列:综合测试 | 9.6 积极术语系列:综合试验

Section outline

Consider the function $\begin{align*}f(x)=\frac{1}{x}\end{align*}$ for $\begin{align*}x\ge1\end{align*}$ . We know that $\begin{align*}\lim\limits_{x \to \infty}\frac{1}{x}=0\end{align*}$ . Is the of $\begin{align*}f(x)\end{align*}$ for $\begin{align*}0 < x < \infty\end{align*}$ , finite? If $\begin{align*}f(x)\end{align*}$ is sampled at the points $\begin{align*}x=n, n=1,2,3,\ldots\end{align*}$ , and these values summed, the series $\begin{align*}\sum\limits_{n=1}^ \infty \frac{1}{n}\end{align*}$ is generated. Do you expect the infinite sum of this series to be finite? The area under the curve question provides the basis for using the Integral Test to help determine convergence or divergence a series: if the area under the curve is finite (infinite) then the infinite summation of discrete values of the associated series must also be finite (infinite).
::将函数 f( x) = 1x 用于 x% 1 。我们知道, limx% 1x= 0 。如果 f( x) 是 0 < x%, 有限吗? 如果 f( x) 在 x=n, n= 1, 2, 3,... 并且这些数值相加, 则生成序列 {n= 11n 。您是否预计此序列的无限总和是有限的? 曲线问题下的区域为使用综合测试来帮助确定趋同或差异提供了依据。如果曲线下的区域是有限的( 无限的) , 那么相关序列的离散值的无限总和也必须是有限的( 无限的) 。

The Integral Test
::综合综合测试

Series with only nonnegative terms, i.e., terms that are either positive or zero, are often called positive term series, and are described as $\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ with $\begin{align*}u_k\ge0\end{align*}$ for every $\begin{align*}k\end{align*}$ . Several of the types of series identified in the previous concepts are, or can be, nonnegative (or positive) term series as shown below:
::仅使用非负值术语的系列,即正值或零值的术语,通常称为正值术语序列,并被描述为+k=1uk,每kk.k.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx/xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx或或或或或或或xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx////或或或或或或或或xx//////////////////////////////

**Common Series Types can be (are) Positive (non-negative) Term**
Series ::系列丛书系列 Type ::类型类型类型	Sigma Notation ::Sigma 符号	Converges if ::组合	Diverges if ::如果	Positive ::阳 Term Series if ::系列(如果)
Arithmetic ::测量学	$\begin{align}S=\sum\limits_{n=1}^\infty[t_0+d(n-1)]\end{align}$	Never ::从未	Always ::总是	$\begin{align}t_0, d\ge 0\end{align}$
Geometric ::几何	$\begin{align}S=\sum\limits_{n=1}^\infty ar^{n-1}\end{align}$	$\begin{align}\|r\|<1\end{align}$ ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ with ::与 $\begin{align}S=\frac{a}{1-r}\end{align}$ ::S=a1-r	$\begin{align}\|r\|\ge1\end{align}$	$\begin{align}a, r \ge 0\end{align}$
Harmonic ::调音	$\begin{align}S=\sum\limits_{n=1}^\infty\frac{1}{n}\end{align}$	Never ::从未	Always ::总是	Always ::总是
$\begin{align}p\end{align}$ -Series ::p- 系列	$\begin{align}S=\sum\limits_{n=1}^\infty\frac{1}{n^p}, p>0\end{align}$ ::Sn=11np,p>0	$\begin{align}p>1\end{align}$	$\begin{align}0 < p \le 1\end{align}$ ::0<p1	Always ::总是

So far we have looked at the following tests for the convergence/divergence of infinite series:
::迄今为止,我们研究了下列无限系列趋同/共振的测试:

Convergence/Divergence Test ::趋同/活力试验	Applicable Series ::适用系列
Limit of $\begin{align}n\end{align}$ th partial sum ::Nn部分金额限额	All ::全部
$\begin{align}n\end{align}$ th-term test for divergence ::nth 差异期度测试	All ::全部

In this and the following two concepts, we will look at convergence tests specifically made for evaluating positive term series:
::在这个概念和以下两个概念中,我们将研究专门为评价正用语系列而作的趋同试验:

The Integral Test
::综合综合测试

Comparison Tests (the Basic, The Simplified Limit Comparison Test)
::比较测试(基础,简化的限制性比较测试)

Ratio and Root Tests
::比率和根测试

This concept will focus on the Integral Test.
::这一概念将侧重于综合测试。

The Integral Test evaluates the integral of the function that looks like the $\begin{align*}n\end{align*}$ th-term of the series. It makes sense to use this kind of test for certain series because the integral is the limit of a certain series.
::集成测试评估函数的内涵, 它看起来像序列的 nth- term。对某些序列使用这种测试是有道理的, 因为集成是某个序列的极限。

The Integral Test states:
::综合测试指出:

Let $\begin{align*}\sum\limits_{k=1}^\infty u_k=\sum\limits_{k=1}^\infty f(k)\end{align*}$ be a series with positive terms.
::@k=1uk_k@k=1f(k) 是一个带有肯定术语的系列。

If $\begin{align*}f(x)\end{align*}$ is a positive, continuous, and decreasing * function for $\begin{align*}x\ge1\end{align*}$ ,
::如果 f(x) 是 x% 1 的正、连续和递减函数 *, 则 f(x) 是一个正、连续和递减函数。

Then:
::然后:

$\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ converges if and only if $\begin{align*}\int\limits_{1}^{\infty}f(x)dx\end{align*}$ converges, i.e. $\begin{align*}\int\limits_{1}^{\infty}f(x)dx=L\end{align*}$ , $\begin{align*}0 < L < \infty\end{align*}$ .
::k=1uk 如果并且只有 {#1f(x)dx 趋同, 也就是 {#1f(x)dx=L, 0<L} 才会趋同。

$\begin{align*}\sum\limits_{k=1}^\infty u_k\end{align*}$ diverges if and only if $\begin{align*}\int\limits_{1}^{\infty} f(x)dx\end{align*}$ diverges, i.e., $\begin{align*}\int\limits_{1}^{\infty} f(x)dx=\infty\end{align*}$ .
::k=1uk 如果且仅在 {#1f(x)dx 出现差异的情况下, 即 {#1f(x)dx} 出现差异, 则会出现差异。

* It may be necessary to determine that $\begin{align*}f^\prime(x)<0\end{align*}$ to see if $\begin{align*}f(x)\end{align*}$ is decreasing everywhere.
::* 可能需要确定f*(x) <0,以确定f(x)是否在下降。

In the statement of the Integral Test, $\begin{align*}u_k\end{align*}$ is assumed to be a function $\begin{align*}f\end{align*}$ of $\begin{align*}k\end{align*}$ . We then change that function $\begin{align*}f\end{align*}$ to be a continuous function of $\begin{align*}x\end{align*}$ in order to evaluate the improper integral of $\begin{align*}f\end{align*}$ . If the improper integral is finite, then the infinite series converges. If the improper integral is infinite, the infinite series diverges. The convergence or divergence of the infinite series depends on the convergence or divergence of the corresponding improper integral.
::在综合测试的语句中, uk 被假定为 k 的函数 f 。然后我们将f 的函数改变为 x 的连续函数, 以便评估 f 的不适当构件。如果不适当的构件是有限的, 那么无限的序列会趋同。如果不适当的构件是无限的, 则无限的序列会不同。无限序列的趋同或差异取决于相应的不适当的构件的趋同或差异。

Let's use the Integral test to show that the $\begin{align*}p\end{align*}$ -series $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{k^5}\end{align*}$ converges.
::让我们用综合体测试来显示 p-series Qk=11k5 的集合。

The series $\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{k^5}\end{align*}$ is a positive term series, with terms that are always defined and are decreasing. The Integral Test can therefore be used to test for convergence or divergence.
::& k=11k5 序列是一个正术语序列, 其术语总是被定义并正在减少。因此, 集成测试可以用来测试趋同或差异。

Let $\begin{align*}f(x)=\frac{1}{x^5}, x\ge 1\end{align*}$ .
::Let f( x) = 1x5, x% 1 。

By the Integral Test, we evaluate $\begin{align*}\int\limits_{1}^{\infty} \frac{1}{x^5}dx\end{align*}$ :
::通过综合测试,我们评估 11x5dx:

$\begin{align*}\int\limits_{1}^{\infty} \frac{1}{x^5}dx & = \lim\limits_{t \to \infty}\int\limits_{1}^{t} \frac{1}{x^5}dx\\ & = \lim\limits_{t \to \infty}\bigg[-\frac{x^{-4}}{4}\bigg]^t_1\\ & = \lim_{t \to \infty}\bigg[-\frac{t^{-4}}{4}+\frac{1}{4}\bigg]\\ & = \frac{1}{4}\end{align*}$
::11x5dx=limttt11x5dx=limt[-x-44]t1=limt[-t-44+14]=14

By the Integral Test we see that the integral is finite and therefore $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{k^5}\end{align*}$ converges.
::通过综合测试,我们看到整体体是有限的,因此Qk=11k5会合。

Now, let's d etermine if the series $\begin{align*}1+ \frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \cdots \frac{1}{4n-3} + \cdots\end{align*}$ converges or diverges using the Integral Test.
::现在,让我们来决定 1+15+19+113+113+14n-3++1+15+19+113+113+113+14n-3++1++15+19+13+13++13+13+13+13+13+13+13+13+++3++++1++15+19+19+19+19+19+19+19+19+13+13+13+13+13+13+13+13+++3++++++13+13+13++13+++13+13+13+13+13++13++13++13+13++13+13+13+13+13+13+13+13+13+13+13+13+13+19+19+19+19+19+19+19+19+19+19+13+13+13+13+13+13+13+13+13+13+13+13++13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13++13+13+13+13+13+++13++++13+13++13+13+13+13+13+++++++++++++++++13++++++++++++++13+13+13+13++++++++++++13+13+13+13+13+19+19+13++++13+13+13++13+13+13+13++++13+++++++++++++13+13+13+13++++++13+13+13++13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13+13++13+13+13+13+++++++++++++19+19+19+13+++++++++++13++++

The series $\begin{align*}1+ \frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \cdots \frac{1}{4n-3} + \cdots\end{align*}$ is a positive term series, with terms that are always defined and are decreasing. The Integral Test can therefore be used to test for convergence or divergence.
::1+15+19+11314n-3是一个积极的术语序列,其术语总是被定义并正在减少。因此,综合测试可用于测试趋同或差异。

Let $\begin{align*}f(x)= \frac{1}{4x-3}, x\ge 1\end{align*}$ .
::Let f( x) =14x - 3, x% 1

By the Integral Test, we evaluate $\begin{align*}\int\limits_{1}^{\infty} \frac{1}{4x-3}dx\end{align*}$ :
::通过综合测试,我们评估 114x -3dx:

$\begin{align*}\int\limits_{1}^{\infty} \frac{1}{4x-3}dx &= \lim\limits_{p \to \infty}\int\limits_{1}^{p} \frac{1}{4x-3}dx\\ & = \lim\limits_{p \to \infty} \bigg[\frac{1}{4}\ln(4x-3)\bigg]^p_1\\ & = \frac{1}{4}\lim\limits_{p \to \infty} [\ln(4p-3)-\ln 0]\\ & = \infty \end{align*}$
::114x- 3dx=limpp114x- 3dx=limp[14ln(4x- 3)]p1=14limp[(4p- 3)- ln0]{_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

By the Integral Test we see that the integral is not finite and therefore the series $\begin{align*}1+ \frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \cdots \frac{1}{4n-3} + \cdots\end{align*}$ diverges.
::通过综合测试,我们发现整体体不是有限的,因此,1+15+19+113+14n-33的系列有差异。

Examples
::实例

Example 1
::例1

The area under the curve of $\begin{align*}f(x)=\frac{1}{x}\end{align*}$ is $\begin{align*}\int\limits_{1}^{\infty}\frac{1}{x}dx=\lim\limits_{m \to \infty}\int\limits_{1}^{m}\frac{1}{x}dx=\lim\limits_{m \to \infty}[\ln m-\ln 1]=\infty\end{align*}$ .
::f(x) = 1x 曲线下的区域是 11xdx = limmmm11xdx =limm[lnm-ln1]。

Although this result might seem counterintuitive because $\begin{align*}\lim\limits_{x \to \infty}\frac{1}{x}=0\end{align*}$ , it suggests that $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{n}\end{align*}$ is not finite, and therefore diverges.
::虽然这个结果可能似乎与直觉相反, 因为 limx=%1x=0, 这表明 {n=11n 并非有限, 因而不同。

Example 2
::例2

Determine if $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{(2k+1)^\frac{3}{2}}\end{align*}$ converges or diverges.
::确定 @k=11( 2k+1) 32 是否相近或有分歧。

The series $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{(2k+1)^\frac{3}{2}}\end{align*}$ is a positive term series, with terms that are always defined and are decreasing. The Integral Test can therefore be used to test for convergence or divergence.
::_k=11( 2k+1) 32 序列是一个正术语序列, 其术语总是被定义并正在减少。因此, 集成测试可用于测试趋同或差异。

Let $\begin{align*}f(x)=\frac{1}{(2x+1)^\frac{3}{2}},x\ge 1\end{align*}$ .
::Let f( x) =1( 2x+1) 32, x% 1

Write and evaluate the integral form:
::编写和评估完整表格:

$\begin{align*}\int\limits_{1}^{\infty}\frac{1}{(2x+1)^\frac{3}{2}}dx = \lim\limits_{t \to \infty}\int\limits_{1}^{t}(2x+1)^{-\frac{3}{2}}dx\end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Use the following $\begin{align*}u\end{align*}$ -substitution to evaluate the integral:
::使用下列替代方法评价集成件:

$\begin{align*}u&=2x+1\\ du&=2dx\end{align*}$
::u=2x+1du=2dx

Then,
::然后,

$\begin{align*}\lim\limits_{t \to \infty}\int\limits_{1}^{t}(2x+1)^{-\frac{3}{2}}dx & = \lim\limits_{t \to \infty}\int\limits_{3}^{2t+1}u^{-\frac{3}{2}}\frac{du}{2}\\ & = \lim\limits_{t \to \infty} \frac{1}{2}\bigg[-2u^{-\frac{1}{2}}\bigg]^{2t+1}_3\\ & = \lim\limits_{t \to \infty}\bigg[-\frac{1}{\sqrt{2t+1}}+ \frac{1}{\sqrt{3}}\bigg]\\ & = \frac{\sqrt{3}}{3}\end{align*}$
::立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方立方

Since the integral is finite, the series $\begin{align*}\sum\limits_{k=1}^\infty \frac{1}{(2k+1)^{\frac{3}{2}}}\end{align*}$ converges by the Integral Test.
::由于整体体是有限的, 序列 *k=11( 2k+1) 32 由综合测试聚合。

Example 3
::例3

The general harmonic series has the form $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{an+b}\end{align*}$ , where $\begin{align*}a\ne 0, b\end{align*}$ are real numbers.
::一般口音序列的窗体为 @n=11an+b,其中a++0,b为真实数字。

Use the Integral Test to show under what conditions the general harmonic series might converge or diverge .
::使用综合测试来显示一般口音序列在什么条件下会趋同或分裂。

To use the Integral Test to determine convergence or divergence of an infinite series:
::利用综合试验确定无限系列的趋同或分歧:

the function $\begin{align*}f(x)=\frac{1}{ax+b}> 0\end{align*}$ for all $\begin{align*}x\ge 1\end{align*}$ ; this means $\begin{align*}ax+b> 0\end{align*}$ for all $\begin{align*}x\end{align*}$ ;
::函数 f( x) = 1ax+b>0 用于全部 x_ 1; 这意味着所有 x 的 ax+b>0 ;

the function $\begin{align*}f(x)=\frac{1}{ax+b}\end{align*}$ must be continuous: $\begin{align*}ax+b\ne 0,x\ge 1\end{align*}$ ; $\begin{align*}a+b>0,b>-a\end{align*}$ ;
::函数 f( x) = 1ax+b 必须连续: ax+b=0, x%1; a+b>0, ba;

the function $\begin{align*}f(x)=\frac{1}{ax+b}\end{align*}$ must be decreasing: $\begin{align*}f^\prime(x)=-\frac{a}{(ax+b)^2}<0\Rightarrow a>0\end{align*}$ .
::函数 f( x) = 1ax+b 必须递减 : fä( x) a( ox+b) 2 < 0a>0 。

If these conditions apply, then the Integral Test can be used:
::如果适用这些条件,可使用综合测试:

$\begin{align*}\int\limits_{1}^{\infty}\frac{1}{ax+b}dx & = \lim\limits_{p \to \infty}\int\limits_{1}^{p}\frac{1}{ax+b}dx\\ & = \lim\limits_{p \to \infty}\bigg[\frac{1}{a}\ln(ax+b)\bigg]^p_1\\ & = \frac{1}{a}\lim\limits_{p \to \infty}\bigg[\ln(ap+b)-\ln(a+b)\bigg]\\ & = \infty\end{align*}$
::= 11ax+bdx=limppp11ax+bdx=limp[1aln(ax+b)]p1=1alimp[ln(ap+b)-ln(a+b)]

By the Integral Test, the general harmonic series $\begin{align*}\sum\limits_{n=1}^\infty \frac{1}{an+b}\end{align*}$ diverges for $\begin{align*}a>0,b>-a\end{align*}$ .
::通过综合测试,普通口音序列= 11an+b的大于0,ba的差异。

Review
::回顾

Maria uses the integral test to determine if $\begin{align*}\sum\limits_{k=1}^\infty\frac{3}{k^2}\end{align*}$ converges. She finds that $\begin{align*}\int\limits_{1}^{+\infty}\frac{3}{x^2}=3\end{align*}$ . She then states that $\begin{align*}\sum\limits_{k=1}^\infty\frac{3}{k^2}\end{align*}$ converges and the sum is 3. What error did she make?
::Maria 使用集成测试来确定 {k=13k2convergets} 是否是k=13k2convergets。她发现 13x2=3。她然后说 k=13k2 相交, 和是 3。她犯了什么错误?

For #2-15, determine if the series converges or diverges using the Integral Test.
::对于# 2-15, 确定该序列是否使用综合测试相交汇或出现差异。

$\begin{align*}\sum\limits_{n=3}^\infty\frac{5}{n-2}\end{align*}$
::#n=35n-2# #n=35n-2#

$\begin{align*}\sum\limits_{n=1}^\infty\frac{n+2}{n+1}\end{align*}$
::*n=1n+2n+1

$\begin{align*}\sum\limits_{n=1}^\infty\frac{n}{3n^2+2}\end{align*}$
::*n=1n3n2+2

$\begin{align*}\frac{1}{3}\ln3+\frac{1}{4}\ln{4}+\frac{1}{5}\ln{5}+\ldots\end{align*}$
::13ln3+14ln4+15ln5+...

$\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{(n+1)\ln(n+1)}\end{align*}$
::*n=11(n+1)x(n+1)

$\begin{align*}\sum\limits_{n=2}^\infty\frac{1}{n(\ln n)^2}\end{align*}$
::*n=21n(lnn)2

$\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{n^{\frac{1}{4}}}\end{align*}$
::*n=11n14 *n=11n14 *n=11n14 *n=11n14 *n=11n14

$\begin{align*}\sum\limits_{k=1}^\infty\frac{7}{\sqrt[5]{k^2}}\end{align*}$
::k=175k2

$\begin{align*}\sum\limits_{n=1}^\infty e^{-n}\end{align*}$
::*n=1e-n

$\begin{align*}\sum\limits_{k=1}^\infty\frac{1}{(3k-1)^{\frac{5}{2}}}\end{align*}$
::*k=11(3k- 1)52

$\begin{align*}\sum\limits_{n=2}^\infty\frac{\ln n}{n^2}\end{align*}$
::*n=2nnn2

$\begin{align*}\sum\limits_{n=1}^\infty\frac{1}{n^2+4}\end{align*}$
::*n=11n2+4

$\begin{align*}\sum\limits_{n=1}^\infty n e^{-n^2}\end{align*}$
::*n=1ne-n2 *n=1ne-n2 *n=1ne-n2 *n=1ne-n2 *n=1ne-n2

$\begin{align*}\sum\limits_{n=1}^\infty\frac{3n+2}{n(n+1)}\end{align*}$
::*n=13n+2n(n+1)

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。

Section outline

The Integral Test ::综合综合测试

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)